Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

A solid circular disc $(m, R)$ translates with $v_{0}$ and rotates with $ω_{0} = \frac{2 v_{0}}{R}$ , at $t = 0$ . Because of friction, pure rolling starts at $t = t_{0}$ . The coefficient of friction is $μ_{k}$ .

Find velocity of centre of disc at $t = t_{0}$ .

Accepted Answer

The frictional force provides the torque required for pure rolling motion(Torque is being calculated around an axis passing through the centre of the disc)

$τ = r \times f \Rightarrow τ = f r \sin (θ)$

here $θ = 90 °$

$τ = f r$

We use the maximum value of friction possible here

$τ = μ_{k} m g r$

Now torques can also be written as $τ = I α$

So,

$- I α = μ_{k} m g r$ .........(1)

Here negative sign represents the retarding nature of the torque due to friction

For a disc, the moment of inertia around an axis passing through its centre is $\frac{m r^{2}}{2}$

Using in equation (1)

$- \frac{m r^{2}}{2} α = μ_{k} m g r \Rightarrow α = \frac{- 2 μ_{k} g}{r}$

Let us assume, the velocity of the centre of disc when pure rolling has begun i.e. at time $t_{o}$ is $v$ .

Along the horizontal direction the only force acting is that of friction, thus:

$m a = f \Rightarrow m a = μ_{k} m g \Rightarrow a = μ_{k} g$

Here $a$ Is constant and hence we can use

$v - u = a t \Rightarrow v - v_{o} = μ_{k} g t_{o} \Rightarrow v = v_{o} + μ_{k} g t_{o}$

Since there is a constant angular acceleration $α = \frac{- 2 μ_{k} g}{r}$

We can use $ω - ω_{o} = α t$

$\Rightarrow \frac{v}{r} - ω_{o} = \frac{- 2 μ_{k} g}{r} t_{o} \Rightarrow \frac{v_{o} + μ_{k} g t_{o}}{r} - \frac{2 v_{o}}{r} = \frac{- 2 μ_{k} g}{r} t_{o} \Rightarrow v_{o} + μ_{k} g t_{o} - 2 v_{o} = - 2 μ_{k} g t_{o} \Rightarrow - v_{o} = - 3 μ_{k} g t_{o} \Rightarrow t_{o} = \frac{v_{o}}{3 μ_{k} g}$

Therefore, the velocity of the centre of the disc at time $t_{o}$ is

$v = v_{o} + μ_{k} g \frac{v_{o}}{3 μ_{k} g} \Rightarrow v = v_{o} + \frac{v_{o}}{3} \Rightarrow v = \frac{4 v_{o}}{3}$

Hence, the velocity of the disc when pure rolling starts is $\frac{4 v_{o}}{3}$

A solid circular disc $(m, R)$ translates with $v_{0}$ and rotates with $ω_{0} = \frac{2 v_{0}}{R}$ , at $t = 0$ . Because of friction, pure rolling starts at $t = t_{0}$ . The coefficient of friction is $μ_{k}$ .

Find velocity of centre of disc at $t = t_{0}$ .

Important Questions on System of Particles and Rotational Motion

Learn and Prepare for any exam you want !

A solid circular disc m,R translates with v0 and rotates with ω0=2v0R, at t=0. Because of friction, pure rolling starts at t=t0. The coefficient of friction is μk. Find velocity of centre of disc at t=t0.

Important Questions on System of Particles and Rotational Motion

Learn and Prepare for any exam you want !

A solid circular disc $(m, R)$ translates with $v_{0}$ and rotates with $ω_{0} = \frac{2 v_{0}}{R}$ , at $t = 0$ . Because of friction, pure rolling starts at $t = t_{0}$ . The coefficient of friction is $μ_{k}$ .

Find velocity of centre of disc at $t = t_{0}$ .