You visited us 1 times! Enjoying our articles? Unlock Full Access!

Rotational Inertia

A plank of ma...

Question

A plank of mass $m_{1}$ with a bar of mass $m_{2}$ placed on it lies on a smooth horizontal plane. A horizontal force growing with time $t$ as $F = a t$ ( $a$ is constant) is applied to the bar. Find how the accelerations of the plank $w_{1}$ and of the bar $w_{2}$ depend on $t$ , if the coefficient of friction between the plank and the bar is equal to $k$ .

Open in App

Solution

Let us write the Newton's second law in projection from along positive x-axis for the plank and the bar
$f r = m_{1} w_{1}$ , $f r = m_{2} w_{2}$ (1)
At the initial moment, $f r$ represents the static friction, and as the force $F$ grows so does the friction force $f r$ , but up to it's limiting value i.e. $f r = f r_{s (m a x)} = k N = k m_{2} g$ .
Unless this value is reached, both bodies moves as a single body with equal acceleration. But as soon as the force $f r$ reaches the limit, the bar starts sliding over the plank i.e. $w_{2} \geq w_{1}$ .
Substituting here the values of $w_{1}$ and $w_{2}$ taken from equation (1) and taking into account that $f_{r} = k m_{2} g$ , we obtain,
$\frac{(a t - k m_{2} g)}{m_{2}} \geq \frac{k m_{2}}{m_{1}} g$ , were the sign $"="$ corresponds to the moment $t = t_{0}$ (say)
Hence, $t_{0} = \frac{k g m_{2} (m_{1} + m_{2})}{a m_{1}}$
If $t \leq t_{0}$ , then $w_{1} = \frac{k m_{2} g}{m_{1}}$ (constant). and
$w_{2} = \frac{(a t - k m_{2} g)}{m_{2}}$

Suggest Corrections

Similar questions

m_{1}

m_{2}

t_{0}

m_{1} = 8 k g

m_{2} = 2 k g

μ = 1 / 5

m_{1} a n d m_{2}

m_{1} = 8 k g

m_{2} = 2 k g

\frac{g}{4}

μ = \frac{1}{5}

m_{1}

m_{2}

F = 30 N

g = 10 {m/s}^{2})

m_{1} = 8 k g

m_{2} = 2 k g

\frac{g}{4}

μ = \frac{1}{5}

m_{1}

m_{2}

F = 30 N

g = 10 {m/s}^{2})

m_{1}

m_{2}

k

m_{1}

F_{m i n} = k g (m_{1} + \frac{m_{2}}{x})

x

Join BYJU'S Learning Program