Question

A sphere is projected up on an inclined plane with a velocity $v_o$ and angular velocity as shown. The coefficient of friction between the sphere and the plane is $\mu =tan\theta$. Find the total time of rise of the sphere.

• A. $\frac{2v_o}{9gsin\theta }$
• B. $\frac{7v_o}{2gsin\theta }$
• C. $\frac{v_o}{gsin\theta }$
• D. $\frac{7v_o}{9gsin\theta }$

Answer 1

The correct option is C $\frac{v_o}{gsin\theta }$


Initially the sphere will slip(no pure rolling) and after sometime the sphere has pure rolling.
Let $t_1$ be the time of no pure rolling and $t_2$ is the time for which the sphere moves in pure rolling.
Using free body diagram of the sphere,
the net forces along the inclined plane is
$F=f_k+mgsin\theta$
$F=\mu N+mgsin\theta$
Substituing $\mu$ and $N$
$F=tan\theta \times mgcos\theta +mgsin\theta$
$=2mgsin\theta$

The acceleration is given by,
$a=\frac{2mgsin\theta }{m}=2gsin\theta$

Net torque, $f_k\times R=I\alpha$

Using $I=\frac{2}{5}m{R}^2$ we get
$\alpha =\frac{5}{2}\frac{gsin\theta }{R}$

At time $t_1$ when the body attains pure rotation let the its velocity be $v$ and angular speed be $\omega$
$v=v_o-at=v_o-2gsin\theta t_1$
$\omega =0+\alpha t_1=\frac{5}{2}\frac{gsin\theta }{R}{t}_1$
Solving, $t_1=\frac{2v_o}{9gsin\theta }$
From here the time required to reach the maximum height is $t_2$
Acceleration of the rolling body is given by
$a=\frac{gsin\theta }{1+\frac{I_{CM}}{m{R}^2}}$
Substituting $I_{CM}$ we get,
$a=\frac{5}{7}gsin\theta$
At time $t_1$, $v=\frac{5v_o}{9}$
Using this,
$t_2=\frac{7}{9}\frac{v_o}{gsin\theta }$
$T=t_1+t_2=\frac{v_o}{gsin\theta }$

For detailed solution watch next video.