First slide

Rolling motion

Question

A sphere pure rolls on a rough inclined plane with initial velocity 2.8 ms^-1 . Find the maximum distance on the inclined plane.

Moderate

A

2.74 m
B

5.48 m
C

1.38 m
D

3.2 m

Solution

Given, initial velocity of sphere, $u = 2.8 {ms}^{- 1}$
Acceleration on the inclined plane,
$a = \frac{g \sin θ}{1 + \frac{K^{2}}{R^{2}}}$ …(i)
where, K and R are the radius of gyration and radius of sphere, respectively.
Moment of inertia of sphere    $I = m K^{2} = \frac{2}{5} m R^{2}$
For sphere,    $\frac{K^{2}}{R^{2}} = \frac{2}{5}$
Putting this value in Eq. 0), we get
$a = \frac{g \sin 30^{\circ}}{1 + \frac{2}{5}}$    $[∵ θ = 30^{\circ} (given)]$
$= \frac{5 g}{14} = \frac{25}{7} {ms}^{- 2}$ $[∵ g = 10 {ms}^{- 2}]$
If s be the maximum distance travelled by the sphere, then at maximum distance, v =0.
$∴$ From third equation of motion, $v^{2} = u^{2} - 2 a s$
$0 = u^{2} - 2 a s$
$\Rightarrow s = \frac{u^{2}}{2 a} = \frac{(2.8)^{2} \times 7}{2 \times 25} = 1.09 m ≃ 1.38 m$

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