A solid sphere of mass $2 k g$ radius $0.5 m$ is rolling with an initial speed of $1 m s^{- 1}$ goes up an inclined plane which makes an angle of $30^{0}$ with the horizontal plane, without slipping. How long will the sphere take to return to the starting point $A$ ?

$0.80 s$
$0.60 s$
$0.52 s$
$0.57 s$

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Solution

The correct option is D
$0.57 s$

Step 1: The given data
Sphere of mass $m = 2 k g$
The radius of the sphere $R = 0.5 m$
Initial speed $u = 1 m / s$
Final speed $= v$
Acceleration due to gravity $= g$
The inclined plane makes an angle $θ = 30^{0}$
Step 2: Formula used:
Using the equation of motion we have,
$v = u + a t \dots (1)$
Using the formula of acceleration for moving without slipping we have,
$a = \frac{g \sin θ}{1 + \frac{k^{2}}{R^{2}}} \dots (2)$ (Where $k$ is Radius of Gyration)
(since, the friction force also applying on the sphere that is why $a \neq g$ )
We know that, Moment of Inertia in terms of the radius of Gyration can be written as,
$I = m k^{2} \dots (3)$ (Where $k$ is Radius of Gyration)
Moment of Inertia of a sphere,
$I = \frac{2}{5} m R^{2}$
Putting the Moment of Inertia of sphere in equation $(3)$ we get,
$\begin{array}{rcl} I & = & m k^{2} \\ = & \frac{2}{5} m R^{2} \end{array}$
$\Rightarrow$ $\frac{k^{2}}{R^{2}} = \frac{2}{5} \dots (4)$
Step 3 : Find the Acceleration:
Therefore, using the equation $(4)$ in equation $(2)$ we get,
$\begin{array}{rcl} a & = & \frac{g \sin θ}{1 + \frac{2}{5}} \\ = & \frac{9.8 \times \sin 30}{1 + \frac{2}{5}} \end{array}$
$\Rightarrow$ $\begin{array}{rcl} a & = & \frac{5 \times 4.9}{7} \end{array}$
$\Rightarrow$ $\begin{array}{rcl} a & = & 3.5 m / s^{2} \end{array}$
Since the sphere is moving up then the equation of motion will be,
$\begin{array}{rcl} v & = & u - a t \\ 0 & = & u - a t \\ t & = & \frac{u}{a} \end{array}$
Step 4 : Find the time of returning to the point $A$ ,
The sphere is going up then coming down to the same initial position, hence the time it will take will be,
$\begin{array}{rcl} T & = & 2 t \\ T & = & \frac{2 u}{a} \\ = & \frac{2 \times 1}{3.5} \\ T & = & 0.57 s \end{array}$
Hence the time of returning is, $T = 0.57 s$
Hence, the correct answer is option (D).

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A solid sphere of mass 2kg radius 0.5mis rolling with an initial speed of 1ms-1 goes up an inclined plane which makes an angle of 300 with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?0.80s0.60s0.52s0.57s