## A thin hollow sphere of mass ‘m’ is completely filled with a liquid of mass ‘m’ when the sphere rolls with a velocity…

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

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

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

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

## A wheel having moment of inertia 2 kg-m^2 about its vertical axis, rotates at the rate of 60 rpm about this axis…

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

## A uniform disk of mass 300 kg is rotating freely about a vertical axis through its centre with constant angular velocity ωo ….

Q: A uniform disk of mass 300 kg is rotating freely about a vertical axis through its centre with constant angular velocity ωo . 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

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

$\displaystyle 150 \omega_0 = 180 \omega$

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

## A sphere rolls without sliding on a rough inclined plane (only mg and contact forces are acting on the body)….

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:

(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 .

## A ring of mass m and radius R rolls on a horizontal rough surface without slipping due to an applied force F ….

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

## Moment of inertia of a uniform quarter disc of radius R and mass M about an axis through its centre of mass ….

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

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

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

## A rod of mass m and length L , pivoted at one of its ends is hanging vertically . A bullet of the same mass moving …..

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 ωM is achieved for x = xM . 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 ,

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

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

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

## 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 ….

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 ,

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

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

As , vc . t = 50 cm

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

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

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

## A football of radius R is kept on a hole of radius r (r < R) made on a plank kept horizontally ....

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 θmax , the football is about to roll , then N2 = 0 and all the forces (mg & N1) must pass through contact point .  Cos(90 – θmax) = r/R

sin θmax = r/R

## 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 …

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 Io . 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 … Click to See Solution :
Ans: (15)
Sol: 