विद्यार्थी मित्रानो या ब्लोग पोस्ट मध्ये आपण Class 12th Physics चा Chapter 1 - Rotational Dynamics चे Exercise Solution पाहणार आहोत. Rotational Dynamics हा खूप महत्वाचा topic आहे. या Solution मुळे तुम्हांला Rotational Dynamics chapter ची Basic Principles लक्षात येतील. त्याचबरोबर अभ्यासात मदत होईल.
Maharashtra Board Class 12th Physics Chapter 1 - Rotational Dynamics Exercise Solution
👉 Exercise 👈
Q.1) Choose the correct option.
i) When seen from below, the blades of a ceiling fan are seen to be revolving anticlockwise and their speed is decreasing. Select correct statement about the directions of its angular velocity and angular acceleration.
(A) Angular velocity upwards, angular acceleration downwards.
(B) Angular velocity downwards, angular acceleration upwards.
(C) Both, angular velocity and angular acceleration, upwards.
(D) Both, angular velocity and angular acceleration, downwards.
Ans :- (B) Angular velocity downwards, angular acceleration upwards.
Explanation :- When observed from below, the blades of a ceiling fan appear to be revolving anticlockwise, indicating a counterclockwise angular velocity. If the speed is decreasing, it means the angular acceleration is opposite to the direction of angular velocity. So, the angular acceleration would be directed upwards, opposing the downward angular velocity.
ii) A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform vertical circular motion, under gravity. Minimum speed of a particle is 5 m/s. Consider
following statements.
P) Maximum speed must be 5√5 m/s.
Q) Difference between maximum and minimum tensions along the string is 60 N.
Select correct option.
(A) Only the statement P is correct.
(B) Only the statement Q is correct.
(C) Both the statements are correct.
(D) Both the statements are incorrect.
Ans :- (B) Only the statement Q is correct.
iii) Select correct statement about the formula (expression) of moment of inertia (M.I.) in terms of mass M of the object and some of its distance parameter/s, such as R, L, etc.
(A) Different objects must have different expressions for their M.I.
(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
(C) Expression for the M.I. for a parallelepiped rotating about the transverse axis passing through its center includes its depth.
(D) Expression for M.I. of a rod and that of a plane sheet is the same about a transverse axis.
Ans :- (B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
iv) In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is √2.5 . Its radius of gyration about a tangent in its plane (in the same unit) must be
(A) √5
(B) 2.5
(C) 2√2.5
(D) √12.5
Ans :- (B) 2.5
v) Consider following cases:
(P) A planet revolving in an elliptical orbit.
(Q) A planet revolving in a circular orbit. Principle of conservation of angular momentum comes in force in which of these?
(A) Only for (P)
(B) Only for (Q)
(C) For both, (P) and (Q)
(D) Neither for (P), nor for (Q)
Ans :- (C) For both, (P) and (Q)
vi) A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, the ratio ”Rotational K.E.: Translational K.E.: Total K.E.” is
(A) 1:1:2
(B) 1:2:3
(C) 1:1:1
(D) 2:1:3
Ans :- (D) 2:1:3
i) Why are curved roads banked?
Answer :- Curved roads are banked to enhance safety and stability for vehicles traveling at high speeds around the curve. When a road is banked, it provides a component of the normal force that acts as the necessary centripetal force to keep the vehicle moving in a circular path. This reduces the reliance on friction between the tires and the road surface to provide the centripetal force, allowing for smoother and safer navigation of the curve. Additionally, banking helps reduce the risk of vehicles skidding or sliding off the road, especially in adverse weather conditions.
ii) Do we need a banked road for a two-wheeler? Explain.
Answer :- Yes, banked roads can benefit two-wheelers, particularly when traveling at high speeds around curves. Banked roads enhance stability, reduce the risk of skidding, and improve handling for two-wheelers by providing an inclined surface that counters centrifugal forces, keeps tires in better contact with the road, and allows for smoother navigation of curves. While not strictly necessary, banked roads significantly enhance safety and performance for two-wheelers, especially in challenging road conditions or when traveling at higher speeds.
iii) On what factors does the frequency of a conical pendulum depends? Is it independent of some factors?
Answer :- The frequency of a conical pendulum, of string length L and semi-vertical angle θ is,
\( \text{n} = \frac{\text{1}}{\text{2} (\pi) } \sqrt{\frac{\text{g}}{\text{cos} (\theta) }} \)
where g is the acceleration due to gravity at the place. From the above expression, we can see that
i) \(n\alpha\sqrt{g}\)
ii) \(n\alpha\frac{1}{\sqrt{L}}\)
iii) \(n\alpha\frac{1}{\sqrt{\cos\theta}}\)
(if θ increases, cos θ decreases and n increases)
(iv) The frequency is independent of the mass of the bob.
iv) Why is it useful to define the radius of gyration?
Answer :- It is useful to define the radius of gyration because it provides a simplified way to characterize the distribution of mass of an object relative to its axis of rotation. This concept is particularly helpful in understanding and calculating moments of inertia, especially for objects with irregular shapes or distributions of mass. By defining the radius of gyration, calculations become simpler, and comparisons between different objects become easier, aiding in both theoretical analysis and experimental measurements of rotational properties.
v) A uniform disc and a hollow right circular cone have the same formula for their M.I., when rotating about their central axes. Why is it so?
Answer :- The moment of inertia (M.I.) formula for a uniform disc and a hollow right circular cone rotating about their central axes is the same because both shapes have the same distribution of mass relative to their axis of rotation, despite their different geometries. This is due to the similarity in their respective radii of gyration (K).
Formula: \( I=\frac{1}{2}mR^{2}\), where m is the mass and R is the radius of gyration.
Q.3) While driving along an unbanked circular road, a two-wheeler rider has to lean with the vertical. Why is it so? With what angle the rider has to lean? Derive the relevant expression. Why such a leaning is not necessary for a four wheeler?
Answer :- When a bicyclist takes a turn along an unbanked road, the force of friction provides the centripetal force, the normal reaction of the road \(\overrightarrow{N}\) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the center of gravity, \(f_{s}.h\), due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \(\overrightarrow{N}\) and the weight \(\overrightarrow{g}.mga=f_{s}.h_{1}\)
Since the force of friction provides the centripetal force,
\(f_{s}=\frac{mv^{2}}{r}\)
If the cyclist leans from the vertical by an angle , the angle between \(\overrightarrow{N}\) and \(\overrightarrow{F}\),
\(\tan\theta=\frac{f_{s}}{N}=\frac{mv^{2}/r}{mg}=\frac{v^{2}}{rg}\)
Hence, the cyclist must lean by an angle \(\theta=\tan^{-1}\frac{v^{2}}{rg}\)
When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the Centre of gravity due to the friction force. But a car is an extended object with four wheels. So, when the inner wheels just get lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight. Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are:
the lateral limiting force of static friction \( \overrightarrow{f_{s}}\) on the wheels - acting along the axis of the wheels and towards the center of the circular path which provides the necessary centripetal force.
the weight mg acting vertically downwards at the centre of gravity (C.G.)
the normal reaction N of the road on the wheels, upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
\( ma_{r}=\frac{mv^{2}}{r}=f_{s}\) .......(1)
In a simplified rigid-body vehicle model, we consider only two parameters - the height h of the C.G. above the ground and the average distance b between the left and right wheels called the track width.
The friction force fs on the wheels produces a torque t that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.
\(\tau_{t}=f_{s}.h=(mv^{2})h\) ..........(2)
When the inner wheel just gets lifted above the ground, the normal reaction N of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque twhich tends to restore the car back on all four wheels.
\(\tau_{t}=mh.\frac{b}{2}\) ........(3)
The car does not topple as long as the restoring torque \(\tau_{t}\), counterbalances the toppling torque \(\tau_{t}\). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by
\(\frac{mv^{2}}{h}=mg.\frac{b}{2}\)
∴ \(V_{max}=\sqrt{\frac{rbg}{2h}}\) ........(4)
Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force \(\overrightarrow{mg}\) passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower center of gravity.
4) Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object.
Answer :- Consider a particle of mass m attached to a string and revolved in a vertical circle of radius r. At every instant of its motion, the particle is acted upon by its weight \(\overrightarrow{mg}\) and the tension \(\overrightarrow{T}\) in the string. The particle may not complete the circle if the string slackens before the particle reaches the top. This requires that the particle must have some minimum speed.
At the top (point A): Let v₁ be the speed of the particle and T₁ the tension in the string. Here, both \(\overrightarrow{T_{1}}\) and weight \(\overrightarrow{mg}\) is vertically downward. Hence, the net force on the particle towards the center O is T₁ + mg, which is the necessary centripetal force.
∴ \(T_{1}+mg=\frac{mv^{2}}{r}\) ......(1)
To find the minimum value of v₁ that the particle must have at the top, we consider the limiting case when the tension T₁ just becomes zero.
∴ \(\frac{mv^{2}}{r}=mg\)
that is, the particle's weight alone is the necessary centripetal force at point A.
∴ \(v_1^2=gr\)
∴ \(v_1=\sqrt{gr}\) …….(2)
At the bottom (point B): Let v₂ be the speed at the bottom. Taking the reference level for zero potential energy to be the bottom of the circle, the particle has only kinetic energy \(\frac{1}{2}mv_2^2\) at the lowest point.
Total energy at the bottom = KE + PE
= \(\frac{1}{2}mv_2^2+0=\frac{1}{2}mv_2^2\) …….(3)
As the particle goes from the bottom to the top of the circle, it rises through a height h = 2r. Therefore, its potential energy at the top is
mgh = mg(2r)
and, from Eq. (2), its minimum kinetic energy there is
\(\frac{1}{2}mv_1^2=\frac{1}{2}mgr\) .........(4)
Minimum total energy at the top = KE + PE
\(=\frac{1}{2}mgr+2mgr\)
\(=\frac{5}{2}mgr\) ........(5)
Assuming that the total energy of the particle is conserved, total energy at the bottom total energy at the top. Then, from Equation (4) and (5),
\(\frac{1}{2}mv_2^2=\frac{5}{2}mgr\)
The minimum speed the particle must have at the lowest position is
\(v_{2}=\sqrt{5gr}\) ........(6)
At the midway (point C): Let v be the speed at point C, so that its kinetic energy is \(\frac{1}{2}mv_3^2\). At C, the particle is at a height r from the bottom of the circle. Therefore, its potential energy at C is mgr. Total energy at C
\(=\frac{1}{2}mv_3^2+mgr\) ........(7)
From the law of conservation of energy, total energy at C = total energy at B
∴ \(\frac{1}{2}mv_3^2+mgr=\frac{5}{2}mgr\)
∴ \(v_3^2=5gr-2gr=3gr\)
∴ The minimum speed the particle must have midway up is
\(v_{3}=\sqrt{3gr}\) ........(8)
In a non-uniform vertical circular motion, e.g., those of a small body attached to a string or the loop-the-loop maneuvers of an aircraft or motorcycle or skateboard, the body must have some minimum speed to reach the top and complete the circle.
In this case, the motion is controlled only by gravity and zero speed at the top is not possible. However, in a controlled vertical circular motion, e.g., those of a small body attached to a rod or the giant wheel (Ferris wheel) ride, the body or the passenger seat can have zero speed at the top, l.e., the motion can be brought to a stop.
∴ \(T_{2}=\frac{mv_2^{2}}{r}+mg\) …….(1)
And total energy at the bottom = KE+PE
\(=\frac{1}{2}mv_2^{2}+0\)
\(=\frac{1}{2}mv_2^{2}\) ………(2)
∴ \(T_{1}=\frac{mv_1^{2}}{r}-mg\) …….(3)
And the total energy at A=KE+PE
\(=\frac{1}{2}mv_1^{2}+mg(2r)\) ………(4)
Then from equation (1) and (3)
\(T_{2}-T_{1}=\frac{mv_2^2}{r}+mg-(\frac{mv_1^2}{r}-mg)\)
\(=\frac{m}{r}(v_2^2-v_1^2)+2mg\) ........(5)
Assuming that the total energy of the body is conserved, the total energy at the bottom=total energy at the top
Then, from equation (2) and (4)
\(\frac{1}{2}mv_2^{2}=\frac{1}{2}mv_1^{2}+mg(2r)\)
∴ \(v_2^2-v_1^2=4gr\) ........(6)
Substituting this in eq. (5)
\(T_{2}-T_{1}=\frac{m}{r}(4gr)+2mg\)
\(=4mg+2mg\)
\(=6mg\)
Therefore, the difference in the tensions in the string at the highest and the lowest points is 6 times the weight of the body.
5) Discuss the necessity of radius of gyration. Define it. On what factors does it depend and it does not depend? Can you locate some similarity between the centre of mass and radius of gyration? What can you infer if a uniform ring and a uniform disc have the same radius of gyration?
Answer :- i) To determine how masses are distributed in a body around a specified axis of rotation. ii) To calculate the moment of inertia for a body of any shape.
Definition : The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.
Factors on which radius of gyration depends:
i) Shape and size of the body.
ii) Position and configuration of axis of rotation.
iii) Distribution of masses in the body with respect to axis of rotation.
iv) Moment of inertia of the body
The radius of gyration does not depend on the mass of the body.
Similarities between centre of mass and radius of gyration:
i) Axis of rotation of a body passing through centre of mass and radius of gyration are parallel to each other.
ii) The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.
The radius of gyration of a thin ring of radius Rr about its transverse symmetry axis is
\(k_{r}=\sqrt{I_{CM}/M_{r}}=\sqrt{R_r^2}=R_{r}\)
The radius of gyration of a thin disc of radius Rd about its transverse symmetry axis is
\(k_{d}=\sqrt{I_{CM}/M_{r}}=\frac{\sqrt{M_{d}R_{d}/2}}{M_{d}}=\frac{1}{\sqrt{2}}R_{d}\)
Given \(k_{r}=k_{d}\)
\(R_{r}=\frac{1}{R_{d}}\) or, equivalently, \(R_{d}=\sqrt{2R_{r}}\).
6) State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.
Answer :- The theorems of parallel axes and perpendicular axes are applicable under the following conditions:
i) Parallel Axes Theorem: Applicable when calculating the moment of inertia of a rigid body about an axis parallel to and at a distance 'd' from the axis passing through the center of mass.
Mathematical expression: \(I_{parallel}=I_{cm}+Md^{2}\)
Where:
\(I_{parallel}\) = moment of inertia about the parallel axis
\(I_{cm}\) = moment of inertia about the center of mass
M = total mass of the object
\(d^{2}\) = distance between the parallel axes
ii) Perpendicular Axes Theorem: Applicable when calculating the moment of inertia of a planar object about two perpendicular axes in the same plane.
Mathematical expression: \(I_{z}=I_{x}+I_{y}\)
Where:
\(I_{z}\) = moment of inertia about the axis perpendicular to the plane containing axes x and y
\(I_{x}\) = moment of inertia about the x-axis
\(I_{y}\) = moment of inertia about the y-axis
7) Derive an expression that relates angular momentum with the angular velocity of a rigid body.
Answer :-
shows a rigid object rotating with a constant angular speed ω about an axis perpendicular to the plane of paper. For theoretical simplification let us consider the object to be consisting of N number of particles of masses \(m_{1},m_{2},...,m_{N}\) at respective perpendicular distances \(r_{1},r_{2},...,r_{N}\) from the axis of rotation. As the object rotates, all these particles perform UCM with same angular speed ω , but with different linear speeds \(v_{1}=r_{1}\omega,v_{2}=r_{2}\omega,v_{N}=r_{N}\omega\).
Linear momentum of the first particle is of magnitude \(p_{1}=m_{1}v_{1}=m_{1}r_{1}\omega\). Its direction is along that of v1. Its angular momentum is thus of magnitude \(L_{1}=p_{1}r_{1}=m_{1}r_1^2\omega,...,L_{N}=m_{N}r_N^2\omega\)
For a rigid body with a fixed axis of rotation, all these angular momenta are directed along the axis of rotation, and this direction can be obtained by using right hand thumb rule. As all of them have the same direction, their magnitudes can be algebraically added. Thus, magnitude of angular momentum of the body is given by
\(L_{1}=m_{1}r_1^2\omega,L_{2}=m_{2}r_2^2\omega,L_{3}=m_{3}r_3^2\omega,...,L_{N}=m_{N}r_N^2\omega\)
\(=(m_{1}r_1^2+m_{2}r_2^2+...+m_{N}r_N^2)\omega=I\omega\)
Where, \(L=m_{1}r_1^2+m_{2}r_2^2+...+m_{N}r_N^2\) is the moment of inertia of the body about the given axis of rotation.
∴ \(L=I\omega\)
8) Obtain an expression relating the torque with angular acceleration for a rigid body.
Answer :- Torque with angular acceleration for a rigid body, we can start with Newton’s second law for rotational motion, which states :
\(\tau=I.\alpha\)
Where, = the torque applied to rigid body,
\(I\) = the moment of inertia of the rigid body,
\(\alpha\) = the angular acceleration of the rigid body.
9) State and explain the principle of conservation of angular momentum. Use a suitable illustration. Do we use it in our daily life? When?
Answer :- The principle of conservation of angular momentum states that the total angular momentum of a closed system remains constant if no external torques act on it. In other words, if the net external torque acting on a system is zero, then the total angular momentum of the system remains constant over time. Angular momentum L is the rotational analog of linear momentum in translational motion. It is defined as the product of the moment of inertia I and the angular velocity of an object:
\[L=I\omega\]
Where:
\(L\) is the angular momentum,
\(I\) is the moment of inertia,
\(\omega\) is the angular velocity.
To illustrate this principle, consider the example of a figure skater spinning on ice. When a figure skater pulls in their arms while spinning, their moment of inertia decreases because the mass is closer to the axis of rotation. According to the law of conservation of angular momentum, since there are no external torques acting on the skater, their angular momentum must remain constant. Therefore, as the moment of inertia decreases, the angular velocity increases to conserve angular momentum, causing the skater to spin faster.
We encounter the principle of conservation of angular momentum in various aspects of daily life, such as:
i) Figure Skating and Gymnastics
ii) Diving
iii) Gyroscopes
iv) Gymnastics
10) Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.
Answer :- Kinetic energy of a rolling body is,
\(E_{total}=E_{translation}+E_{rotational}\) ..........(1)
Where, \(E_{translation}\) and \(E_{rotational}\) are the kinetic energies associated with the translation of center of mass and rotation about an axis.
Suppose M and R be the mass and radius of the body, , K and I be the angular speed, radius of gyration and moment of inertia for rotation about an axis through its center and v be the translation speed of CV(center of mass).
\(v=\omega R\) and \(I=MK^{2}\) ..........(2)
\(E_{translation}=\frac{1}{2}Mv^{2}\) and \(E_{rotational}=\frac{1}{2}I\omega^{2}\)
\(E=\frac{1}{2}Mv^{2}+\frac{1}{2}I\omega^{2}\)
\(=\frac{1}{2}Mv^{2}+\frac{1}{2}I\frac{v^{2}}{r^{2}}\)
\(=\frac{1}{2}Mv^{2}(1+\frac{1}{MR^{2}})\)
\(=\frac{1}{2}Mv^{2}(1+\frac{MK^{2}}{MR^{2}})\)
∴ \(E=\frac{1}{2}Mv^{2}(1+\frac{K^{2}}{R^{2}})\)
11) A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
Answer :-
Height = h , rolling down distance = L, base = R
Conserving energy,
\((KE+U)_{Initial}=(KE+U)_{Final}\)
\(0+mg=\frac{1}{2}mv^{2}(1+\frac{k^{2}}{r^{2}})+0\)
\(v=\sqrt{\frac{2gh}{(1+\frac{k^{2}}{r^{2}})}}\)
When a = constant, \(v^{2}=u^{2}+2as\)
\(\frac{2gh}{(1+\frac{k^{2}}{r^{2}})}=0^{2}+2a.L\) ......... put S=L
\(a=\frac{gh}{L(1+\frac{k^{2}}{r^{2}})}\)
\(\sin\theta=h/L\) where \(L=h/\sin\theta\)
\(a=\frac{gh}{L(1+\frac{k^{2}}{r^{2}})}\)
\(=\frac{gh}{\frac{h}{\sin\theta}(1+\frac{k^{2}}{r^{2}})}\) , h gets cancel
\(a=\frac{g\sin\theta}{(1+\frac{k^{2}}{r^{2}})}\)
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