Rotational Motion

Q: A thin hollow sphere of mass ‘m’ is completely filled with a liquid of mass ‘m’ when the sphere rolls with a velocity ‘v’ kinetic energy of the system is (neglect friction)

Sol: Total energy = KE + rotational KE

$\large = \frac{1}{2}(2m ) v^2 + \frac{1}{2}(\frac{2}{3}mr^2 ) \omega^2 $

$\large = \frac{1}{2}(2m ) v^2 + \frac{1}{3} m v^2 $

$\large = \frac{4}{3}m v^2 $

Q: A wheel having moment of inertia 2 kg-m^2 about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel’s rotation in one minute would be

(a) 2π/15 N-m

(b) π/12 N-m

(c) π/15 N-m

(d) π/18 N-m

Ans: (c)

Sol: L = I ω  ; Where I = Moment of Inertia

L = I (2 π ν) ; Where ν = frequency

L = 2 (2 π × 1 ) = 4 π

Torque = L/Δt = 4π /60 = π /15 N-m

Q: A uniform disk of mass 300 kg is rotating freely about a vertical axis through its centre with constant angular velocity ω_o . A boy of mass 30 kg starts from the centre and moves along a radius to the edge of the disk. The angular velocity of the disk now is

(A) $\frac{\omega_0}{6}$

(B) $\frac{\omega_0}{5}$

(C) $\frac{4\omega_0}{5}$

(D) $\frac{5 \omega_0}{6}$

Ans: (D)

Solution: As Torque = 0 ; angular momentum remains conserved

$\displaystyle L = (0 + \frac{300R^2}{2})\omega_0 = (\frac{300R^2}{2} + 30 R^2)\omega $

$\displaystyle 150 \omega_0 = 180 \omega $

$\displaystyle \omega = \frac{5}{6}\omega_0 $

Q: A sphere rolls without sliding on a rough inclined plane (only mg and contact forces are acting on the body). The angular momentum of the body:

(A) about centre is conserved

(B) is conserved about the point of contact

(C) is conserved about a point whose distance from the inclined plane is greater than the radius of the sphere

(D) is not conserved about any point

Ans: (C)
Solution: Angular momentum will be conserved if the net
torque is zero . Now for the sphere to move down:
mg sinθ > μ mg cosθ

Let x be the perpendicular distance of the point
(as shown in figure) about which torque remains
zero.
for τ = 0 ; x > R as shown

Note: As mgsinθ > μ mgcosθ , the point should be inside the sphere .

Q: A ring of mass m and radius R rolls on a horizontal rough surface without slipping due to an applied force F . The friction force acting on ring is :

(A) F/3

(B) 2F/3

(C) F/4

(D) Zero

Ans: (D)

Solution: F + f = m a …. (1)

Also ; F R – f R = I a/ R

F – f = m a …. (2)

[I = m R²]

From (1) & (2)

f = 0

Q: Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of mass and perpendicular to its plane is :

(A) $\frac{MR^2}{2} – M(\frac{4R}{3\pi})^2$

(B) $\frac{MR^2}{2} – M(\sqrt{2} \frac{4R}{3\pi})^2$

(C) $\frac{MR^2}{2} + M(\frac{4R}{3\pi})^2$

(D) $\frac{MR^2}{2} + M(\sqrt{2} \frac{4R}{3\pi})^2$

Ans: (B)

Solution:

Moment of Inertia about ‘O’ is $I = \frac{MR^2}{2}$

By parallel-axis theorem

$\frac{MR^2}{2} = I_{cm} + M(\sqrt{2} \frac{4R}{3\pi})^2 $

$I_{cm} = \frac{MR^2}{2} – M(\sqrt{2} \frac{4R}{3\pi})^2 $

Q: A rod of mass m and length L , pivoted at one of its ends is hanging vertically . A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it . The combined system now rotates with angular speed ω about the pivot . The maximum angular speed ω_M is achieved for x = x_M . Then

(a) $\omega = \frac{3 v x}{L^2 + 3 x^2}$

(b) $\omega = \frac{12 v x}{L^2 + 12 x^2}$

(c) $x_M = \frac{L}{\sqrt{3}}$

(c) $\omega_M = \frac{v}{2L} \sqrt{3}$

Ans: (a,c,d)

Solution: According to conservation of angular momentum about point of suspension ,

$\displaystyle m v x = (\frac{m l^2}{3} + m x^2 )\omega$

$\displaystyle v x = (\frac{ l^2}{3} + x^2 )\omega$

$\displaystyle \omega = \frac{3 v x}{L^2 + 3 x^2} $

For maximum ω ,

$\displaystyle \frac{d\omega}{dx} = 0 $

$\displaystyle x = \frac{L}{\sqrt{3}} $

$\displaystyle \omega = \frac{\sqrt{3} v}{2L} $

Q: A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on left) . It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the axle on top of the axle (see figure on right) . The scale is now pushed slowly on the axle so that it moves without slipping on the axle , and the roller starts rolling without slipping . After the roller has moved 50 cm , the position of the scale will look like

Ans: (b)

Solution: For no slipping at ground ,

v_c = ω R ; where r is the radius of roller

Velocity of scale = (v_c + ω r ) ; where r is the radius of axle

As , v_c . t = 50 cm

Distance moved by scale = (v_c + ω r ) t

= (v_c + v_c r/R) t = (3 v_c/2) .t = 75 cm

Therefore , relative displacement with respect to center of roller is = 75 -50 = 25 cm

Q: A football of radius R is kept on a hole of radius r (r < R) made on a plank kept horizontally . One end of plank is now lifted so that it gets tilted making an angle θ from horizontal as shown in the figure below . The maximum value of θ so that the football does not start rolling down the plank satisfies

(a) sin θ = r/R

(b) tan θ = r/R

(c) sin θ = r/2R

(d) cos θ = r/2R

Ans: (a)

Solution: For θ_max , the football is about to roll , then N₂ = 0 and all the forces (mg & N₁) must pass through contact point .

Cos(90 – θ_max) = r/R

sin θ_max = r/R

Q: ABC is a plane lamina of the shape of an equilateral triangle . D, E are the mid points AB , AC and G is the centroid of the lamina . Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I_o . If part ADE is removed the moment of inertia of the remaining part about the same axis is $\frac{N I_o}{16}$ where N is an integer . The value of N is …

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Ans: (15)

Sol: