LinkedIn Insight Solutions for Exercises of Rotational Dynamics - Grad Plus

Solutions for Exercises of Rotational Dynamics

Use g = 10 m/s2, unless, otherwise stated.

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 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.

Answer: We have defined $latex \int\sin\left(\phi\right)d\phi=\frac12$ and we cannot use

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 $latex 5\sqrt5$ 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.


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 centre includes its depth.

(D) Expression for M.I. of a rod and that of a plane sheet is the same about a transverse axis.

iv) In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is $latex \sqrt{2.5}$. Its radius of gyration about a tangent in its plane (in the same unit) must be

$latex (A)\;\sqrt5$

(B) 2.5

$latex (C)\;\:2\sqrt5$

$latex \left(D\right)\;\sqrt{12.5}$

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)

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

2. Answer in brief.

i) Why are curved roads banked?

Along the curve, as the speed of the vehicle increases the magnitude of centrifugal force increases & accident may occur. To avoid this, the curved roads are so constructed that its outer edge is raised to a certain level than the inner edge.

ii) Do we need a banked road for a two-wheeler? Explain.

iii) On what factors does the frequency of a conical pendulum depends? Is it independent of some factors?

iv) Why is it useful to define radius of gyration?

The radius of gyration of a body depends upon, (a) Mass of the body and (b) Distribution of mass about the axis of rotation. Thus radius of gyration (K) serves as an indication of the distribution of mass about the axis of rotation.

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?

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?

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.

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?

6. State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.

7. Derive the expression that relates angular momentum with angular velocity of rigid body.

If a body performing rotational motion is having linear momentum $latex \overrightarrow P$ and position vector from the axis of rotation as $latex \overrightarrow r$ then angular momentum is given as,

$latex \overset\rightharpoonup L=\overrightarrow r\times\overrightarrow P$

Consider a rigid body rotating about an axis passing through a point O with a constant angular velocity ω. Let it be consists of  n  particles of masses m1,m2,m3, m4…….mn at distances r1, r2, r3…..rn respectively from axis of rotation. As the body is rigid, each particle rotates with same angular velocity ω.

Linear velocity of first particle is given as, $latex v_1=r_1\omega$.

And its linear momentum is given by, $latex P_1=m_1v_1=m_1r_1\omega$.

The moment of linear momentum is called angular momentum.

Thus angular momentum of first particle is given by,

$latex L_1=P_1r_1=\left(m_1r_1\omega\right)r_1=m_1r_1^2\omega$.

Similarly for other particles, we have,

$latex L_2=m_2r_2^2\omega,\;L_3=m_3r_3^2\omega,\;…L_n=m_nr_n^2\omega$

Hence angular momentum of rigid body is,

$latex L=L_1+L_2+L_3+……+L_n$

$latex \therefore L=m_1r_1^2\omega+m_2r_2^2\omega+m_3r_3^2\omega+……..+m_nr_n^2\omega$

$latex \therefore L=\left(m_1r_1^2+m_2r_2^2+m_3r_3^2+……..+m_nr_n^2\right)\omega$

$latex \therefore L=\left[\overset n{\underset{i=1}{\sum m_ir_i^2}}\right]\omega=I\omega$

$latex \therefore L=I\omega$

8. Obtain an expression relating the torque with the angular acceleration for a rigid body.

9. State and explain the principle of conservation of angular momentum. Use a suitable illustration. Do we use it in our our daily life? When?

10. Discuss the interlink between translational, rotational and total kinetic energies of a rigid objects that rolls without slipping.

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.

12. Somehow, an ant is stuck to the rim of a bicycle wheel of diameter 1 m. While the bicycle is on a central stand, the wheel is set into rotation and it attains the frequency of 2 rev/s in 10 seconds, with uniform angular acceleration. Calculate (i) Number of revolutions completed by the ant in these 10 seconds. (ii) Time taken by it for first complete revolution and the last complete revolution.

D = 1 m, r = 0.5 m,

n1 = 0, n2 = 2 Hz, t = 10 sec,

tfirst = ?  tlast= ?  N = ?

We know that, $latex \alpha=\frac{2\pi\left(n_2-n_1\right)}t$

$latex \therefore\alpha=\frac{2\pi\times(2-0)}{10}=\frac{4\pi}{10}rad/s^2$

We have, angular displacement,

$latex \vartheta=\omega_1t+\frac12\alpha t^2$

$latex \therefore\theta=0+\frac12\times\frac{4\pi}{10}\times10^2$

$latex \therefore\theta=\frac{2\pi}{10}\times100=2\pi$

Hence number of revolutions completed in these 10 seconds is given as,

$latex N=\frac\theta{2\pi}=\frac{20}{2\pi}=10$

For first revolution,

θ = 2π and ω1 = 0

$latex \therefore\;u\sin g\;,\;\theta=\omega_1t+\frac12\alpha t^2$

we have, $latex 2\pi=0+\frac12\left(\frac{4\pi}{10}\right)t^2$

$latex \therefore\;t^2=\frac{2\pi\times20}{4\pi}=10$

$latex \therefore\;t_{first}=\sqrt{10}st_{first}=\sqrt{10}s$

Now let us find time required to complete 9 revolutions.

We have,

θ = 9 × 2π = 18π and ω1=0

$latex \therefore\;u\sin g\;\theta=\omega_1t+\frac12\alpha t^2$

we have, $latex 18=0+\frac12\left(\frac{4\pi}{10}\right)t^2$

$latex \therefore\;t^2=\frac{18\pi\times20}{4\pi}=90$

$latex \therefore\;t_9=\sqrt{90}=9.486\;s$

It is given that time required for 10 complete revolutions is 10s.

Hence time required to complete 10th revolution is given as,

$latex \therefore\;t_{last}=t_{10}-t_9$

t = 10 − 9.486

t = 0.514s

13. Coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the centre of the coin is 7c cm away from the centre of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim? (use g = π2 m/s2)

µ = 0.5,                    R = 8cm = 8 × 10-2m,

r = πcm=π×10-2,  n=? for slipping of coin,

We know that for horizontal circular motion,

Maximum velocity is given as, $latex V_{max}=\sqrt{\mu rg}$

$latex \therefore V_{max}=\sqrt{\mu rg}$

$latex =\sqrt{0.5\times\pi\times10^{-2}\times\pi^2}$

But, v = rω

i.e. v = 2πnr

$latex \therefore\;n_{max}\;=\;\frac{V_{max}}{2\pi r}$

$latex \therefore n_{max}=\frac{\sqrt{0.5\times\pi\times10^{-2}\times\pi^2}}{2\times\pi\times\pi\times10^{-2}}$

$latex =\frac{10^{-1}\times\pi\sqrt{0.5\times\pi}}{2\times\pi\times\pi\times10^{-2}}$

$latex =\frac{\sqrt{0.5\times\pi}}{2\times\pi\times10^{-1}}$

$latex \therefore\;n_{max}\propto\frac1{\sqrt r}$

$latex \therefore\;\frac{(n_{max})R}{(n_{max})r}\;\sqrt{\frac r$

$latex =\sqrt{\frac{\pi\times10^{-2}}{8\times10^{-2}}}$

$latex =0.6264$

Hence, if coin is placed at the rim answer will change by the factor 0.6264.

14. Part of a racing track is to be designed for a curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle should the road be tilted? By what height will its outer edge be, with respect to the inner edge if the track is 10 m wide?

r = 72 m, $latex v\;=\;216kmph=216\times\frac5{18}=60\;m/s,$

θ = ?, H = ? If L = 10m

We know that,

$latex \theta=\tan^{-1}\left(\frac{v^2}{rg}\right)$

$latex \theta\;=\;\tan^{-1}\left(\frac{60\times60}{72\times9.8}\right)$

$latex \theta\;=\;\tan^{-1}(5.10)\;=\;9.822$

Now, h = Lsinθ

∴ h = 10 × sin(78.69°)

∴ h = 10 × 0.9822

∴ h = 9.822

∴ tan-1(5.1)=78.69º

h=9.8 m

15. The road in the question 14 above is constructed as per the requirements. The coefficient of static friction between the tyres of a vehicle on this road is 0.8, will there be any lower speed limit? By how much can the upper speed limit exceed in this case?

µ = 0.8, g = 9.8m/s

The minimum speed limit is imposed by the frictional force
component and we know that, $latex V_{min}=\sqrt{\mu rg}$

$latex V_{min}\;=\;\sqrt{0.8\times72\times9.8}$

$latex V_{min}\;=\;\sqrt{564.48}$

$latex V_{min}\;=\;\sqrt{5.64\times10^2}$

$latex V_{min}\;=\;\sqrt{5.6448}\;\times\;10$

$latex V_{min}\;=\;23.76\;m/sec$

$latex V_{min}\;=\;24\;\times\;\frac5{18}\;km/h$

$latex V_{min}=86.4\;km/h$

As road is banked with θ > 45° limit for the speed.

16. During a stunt, a cyclist (considered to be a particle) is undertaking horizontal circles inside a cylindrical well of radius 6.05 m. If the necessary friction coefficient is 0.5, how much minimum speed should the stunt artist maintain? Mass of the artist is 50 kg. If she/he increases the speed by 20%, how much will the force of friction be?

r = 6.05m,    µ = 0.5, vmin = ?,

m = 50 kg,   Frictional force = ? if speed in increased by 20%

We know that for cylindrical well, $latex V_{min}=\sqrt{\frac{rg}\mu}$

$latex \therefore V_{min}=\sqrt{\frac{rg}\mu}$

$latex =\;\sqrt{\frac{6.05\times10}{0.5}}\;=\;11m/s$

In the well of depth, force of friction is equal to weight of the object i.e. f = mg

∴ f = 50 × 10

i.e. f = 500N

17. A pendulum consisting of a massless string of length 20 cm and a tiny bob of mass 100 g is set up as a conical Its bob now performs 75 rpm. Calculate kinetic energy and increase in the gravitational potential energy of the bob. (Use π2 =10 )

m =0.1 kg, ???? = 20cm = 20 × 10-2m,

$latex n\;=\;\frac{75}{60}RPS\;=\;\frac54Hz,$

$latex K.E=?,\;\triangle U=?$

We know that,

h = ????cosθ and r = ????sinθ

$latex Also,\;T=2\pi\sqrt{\frac{l\cos\theta}g}$

$latex \therefore\;T=\frac1n=2\pi\sqrt{\frac{l\cos\theta}g}$

$latex \therefore\;\frac45=2\pi\sqrt{\frac{0.2\cos\theta}{10}}$

$latex \therefore\;\frac{16}{2s}=\frac{4\pi^20.2\cos\theta}{10}$

$latex \therefore\;\cos\theta=\frac{16\times20}{25\times4\times10\times0.2}$

$latex \therefore\;\cos\theta\;=\;0.8$

∴ θ = 37°

Now, we know that,

⇒v2 = rgtanθ

∴ v2=1sin37º gtan37º

$latex =0.2\times\frac35\times10\times\frac34$

$latex v^{{}^2}=\frac9{10}$

$latex \therefore\frac12mv^2=\frac12\times0.1\times\frac9{10}$

$latex \therefore K.E=\frac12mv^2=0.045\;Joule$

$latex \triangle U=mgd=mg(l-h)=mg(l-l\cos\theta)$

$latex \therefore\;\triangle U\;=\;mgl(1-\cos\vartheta)\\$

$latex =\;0.1\;\times\;10\;\times\;0.2(1-0.8)\\$

$latex \therefore\;\triangle U\;=\;10\;\times\;0.2\;\times\;0.2\\$

$latex =0.1\times10\times0.2(1-0.8)\\$

18. A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate diameter of the sphere of death. How much minimum values are possible for these two speeds?

19. A metallic ring of mass 1 kg has moment of inertia 1 kg m2 when rotating about one of its diameters. It is molten and remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis.

m = 1kg, I = 1 kgm2

Molten and converted into disc then I’ = ?

$latex For\;ring.\;About\;its\;diameter\;,I=\frac{mr^2}2\\$

$latex \therefore1=\frac{1\times r^2}2\\$

$latex \therefore r=2\;\;\;\;\;\;\;\;\;\;\;\;\;\therefore r=\sqrt2m\\$

Now after converting into the disc of same radius,

$latex I’\;=\;\frac{mr^2}2\;=\;\frac{mr^2}2\\$

$latex =\;\frac{1\times{(\sqrt2)}^2}2\;=\;1kgm^2\\$

20. A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid spheres of mass 50 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumb-bell when rotated about an axis passing through its centre and perpendicular to the length.

mR= 60g = 60 × 10-3kg,????=20cm=20×10-2m

mS= 50g = 50 × 10-3kg, r=10cm=10×10-2m

$latex I_s=m_s\times\left(\frac l2\right)^2\\$

$latex I_s=50\times10^{-3}\times\left(\frac{20\times10^{-2}}2\right)^2\\$

$latex I_s\;=\;50\;\times\;10^{-3\;}\times10^{-2}\\$

$latex \therefore I_s=50\times10^{-5}kgm^2\\$

$latex \therefore\;I_r\;=\;\frac{mrl^2}{12}$

$latex \therefore\;I_r\;=\frac1{12}\times60\times10^{-3}\times{(20\times10^{-2})}^2$

$latex \therefore I_r=0.2\times10^{-5}kgm^2$

Hence total moment of inertia is given as,

Is = (2 × Is)+ Ir = (100 × 10-5)+(1.2×10-5)


i.e. Is=10020 gcm2

21. A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10 kg and radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain distance x from the centre. Calculate x if the wheel now rotates at 80 rpm.

ω1=100rpm=100/60rps = 5/3 rps, m1=10kg

r = 0.4 m, m2 = 1.6 kg at a distance of x from centre.

x=? if ω2=80rpm = 80/60rps = 4/3rps

Moment of inertia of the disc before sticking clay,


$latex =\;\frac{10\times0.4\times0.4}2\;=\;0.8kgm^2$

Moment of inertia after sticking the clay at distance of x from the centre,

$latex I_2=I_1+m_2x^2$

$latex \therefore\;I_2=0.8+1.6x^2$

According to law of conservation of momentum,

$latex I_1\omega_1=I_2\omega_2$

$latex \therefore\;0.8\times\frac53=(0.8+1.6x^2)\times\frac43$

$latex \therefore\;(0.8+1.6x^2)=1$

$latex \therefore\;1.6x^2=0.2$

$latex \therefore\;x^2=\frac{0.2}{1.6}=\frac18$

$latex \therefore\;x=\frac1{2\sqrt2}=0.35m$

22. Starting from rest, an object rolls down along an incline that rises by 3 in every 5 (along it). The object gains a speed of $latex \sqrt{10}\;m/s$ as it travels a distance of 5⁄3m along the incline. What can be the possible shape/s of the object?

Incline rises by 3 for every 5 hence,

$latex \sin\theta=\frac35$

After distance s=5/3 it gains speed $latex v=\sqrt{10}$, u=0

We know the relation between linear distance along the incline and the vertical distance (height) given as as,

$latex \sin\theta=\frac hs$

$latex \frac35=\frac h{\displaystyle5/3}\;\;\;\;\;\;\;\;\;\;\;\therefore\;h=1m$

We know that velocity for the rolling object is given

$latex v\;=\;\sqrt{\frac{2gh}{1+{\displaystyle K^2/}R^2}}$

$latex \;\therefore v^2=\frac{2gh}{1+{\displaystyle K^2/R^2}}$

$latex \therefore\;10\;=\;\frac{\displaystyle2\times10\times1}{\displaystyle1+K^2/R^2}\;\;\;\;\;as\;\frac{\displaystyle K^2}{\displaystyle R^2}=1$

$latex As\;\frac{K^2}{R^2}=1,\;It\;can\;be\;a\;ring\;or\;a\;hollow\;cylinder.$

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