Question
A tangential force of $$20\,N$$ is applied on a cylinder of mass $$4\,kg$$ and moment of inertia $$0.02\,kg\,{m^2}$$ about its own axis. If the cylinder rolls without slipping, then linear acceleration of its centre of mass will be
A tangential force of $$20\,N$$ is applied on a cylinder of mass $$4\,kg$$ and moment of inertia $$0.02\,kg\,{m^2}$$ about its own axis. If the cylinder rolls without slipping, then linear acceleration of its centre of mass will be
A.
$$6.7\,m/{s^2}$$
B.
$$10\,m/{s^2}$$
C.
$$3.3\,m/{s^2}$$
D.
None of these
Answer :
$$6.7\,m/{s^2}$$
Solution :
Let friction force $$= f$$
$$\eqalign{ & F + f = ma\,......\left( {\text{i}} \right) \cr & \left( {F - f} \right)R = I\alpha \,......\left( {{\text{ii}}} \right) \cr} $$
From eqns. (i) and (ii),
$$2F = ma + \frac{{I\alpha }}{R}$$
Use $$\alpha = \frac{a}{R}\left( {{\text{for pure rolling}}} \right)$$
$$\eqalign{ & 2F = ma + \frac{{Ia}}{{{R^2}}} \cr & 40 = 4a + \frac{{0.02a}}{{{{\left( {0.1} \right)}^2}}};a = \frac{{40}}{6} = 6.7\,m/{s^2} \cr} $$
Let friction force $$= f$$
$$\eqalign{ & F + f = ma\,......\left( {\text{i}} \right) \cr & \left( {F - f} \right)R = I\alpha \,......\left( {{\text{ii}}} \right) \cr} $$
From eqns. (i) and (ii),
$$2F = ma + \frac{{I\alpha }}{R}$$
Use $$\alpha = \frac{a}{R}\left( {{\text{for pure rolling}}} \right)$$
$$\eqalign{ & 2F = ma + \frac{{Ia}}{{{R^2}}} \cr & 40 = 4a + \frac{{0.02a}}{{{{\left( {0.1} \right)}^2}}};a = \frac{{40}}{6} = 6.7\,m/{s^2} \cr} $$
