A round uniform body of radius ‘R’ , mass ‘M’ and moment of inertia ‘I’, rolls down (without slipping)and inclined plane…

Q: A round uniform body of radius ‘R’ , mass ‘M’ and moment of inertia ‘I’, rolls down (without slipping)and inclined plane making an angle ‘θ’ with the horizontal. Then, its acceleration is

Sol: $\large v = \sqrt{\frac{2 g h}{1+\frac{I}{MR^2}}}$

$\large h = L sin\theta$

$\large v^2 = u^2 – 2 a s$

$\large a = \frac{v^2}{2 s} = \frac{2 g L sin\theta}{2(1+\frac{I}{MR^2})L}$

$\large a = \frac{g sin\theta}{(1+\frac{I}{MR^2})}$

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 pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t^2) N (where, t is measured in seconds) applied tangentially…

Q: A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t2) N (where, t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , then the number of rotations made by the pulley before its direction of motion is reversed, is

Sol: Given force, F = 20t – 5t2

$\large \alpha = \frac{F R}{\alpha}$

$\large \alpha = \frac{(20 t – 5t^2)2}{10}$

= 4 t – t2

$\large \frac{d\omega}{dt} = 4 t – t^2$

$\large \int_{0}^{\omega} d\omega = \int_{0}{t} (4 t – t^2 ) dt$

$\large \omega = 2 t^2 – \frac{t^3}{3}$

When direction is reversed

ω = 0, i.e., t = 0 to 6s Now,

$\large d\theta = \omega dt$

$\large \int_{0}^{\theta} d\theta = \int_{0}^{6}(2 t^2 – \frac{t^3}{3}) dt$

$\large \theta = [\frac{2t^3}{3} – \frac{t^4}{12}]_{0}^{6}$

θ = 144 – 108 = 36 rad

Number of rotations, $\large n = \frac{\theta}{2\pi} = \frac{36}{2\pi} < 6$

A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end…

Q: A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω. Its centre of mass rises to a maximum height of

Sol: $\large m g h = \frac{1}{2}I \omega^2$

$\large m g h = \frac{1}{2} (\frac{m l^2}{3}) \omega^2$

$\large h = \frac{l^2 \omega^2}{6 g}$

A wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle…

Q: A wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of 15° in time t sec. Then how much it rotates in the next 2t sec.

Sol: If angular acceleration is constant, we have

$\large \theta = \omega_0 t + \frac{1}{2}\alpha t^2$

$\large 15^o = \frac{1}{2}\alpha t^2$ …(i)

for the second condition (time = 3t sec)

$\large \theta’ = \frac{1}{2}\alpha (3t)^2$

$\large \theta’ = 9 \times 15 = 135^o$ …(ii)

$\large \Delta \theta = \theta’ – \theta$

$\large \Delta \theta = 135^o – 15^o = 120^o$

Moment of inertia of a body about an axis is 4 kg m2. The body is initially at rest and a torque of 8 Nm starts acting on it along the same axis…

Q: Moment of inertia of a body about an axis is 4 kg m2. The body is initially at rest and a torque of 8 Nm starts acting on it along the same axis. Work done by the torque in 20 s, in joules is

Sol: $\large \tau = I \alpha$

$\large \alpha = \frac{\tau}{I}$

$\large \alpha = \frac{8}{4} = 2$

$\large \theta = \frac{1}{2} \alpha t^2$

$\large \theta = \frac{1}{2} \times 2 \times (20)^2$

= 400

$\large W = \tau \theta$

$\large W = 8 \times 400$

= 3200 J

A fly – wheel of mass 25kg has a radius of 0.2m. It is making 240 rpm. What is the torque necessary to bring to rest in 20 s ?

Q: A fly – wheel of mass 25kg has a radius of 0.2m. It is making 240 rpm. What is the torque necessary to bring to rest in 20 s ?

Sol: $\large \alpha = \frac{2\pi \nu}{t}$

$\large \alpha = \frac{2\pi \times 4}{20}$

Torque , $\large \tau = I \alpha$

$\large \tau = \frac{MR^2}{2} \alpha$

= 0.2π Nm