First slide

Rolling motion

Question

A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle inclination 30⁰. The centre of mass of cylinder has speed of 4 ms^-1. The distance travelled by the cylinder on the inclined surface will be
(Take, g = 10 ms^-2

Moderate

A

2.2 m
B

1.6 m
C

1.2 m
D

2.4 m

Solution

When a body rolls, i.e. have rotational motion, the total kinetic energy of the system will be
$KE = \frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})$
where, m = mass of body, v = velocity and K = radius of gyration.
Given, $m = 2 kg, θ = 30^{\circ} and v = 4 {ms}^{- 1}$
Let h be the height of the inclined plane, then from law of conservation of energy,
KE=PE
$\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}}) = m g h$
Substituting the given values in the above equation, we get
$\frac{1}{2} \times 2 \times 16 (1 + \frac{1}{2}) = 2 \times 10 \times h (For cylinder, \frac{K^{2}}{R^{2}} = \frac{1}{2})$
$\Rightarrow 8 \times \frac{3}{2} = 10 h \Rightarrow h = 1.2 m$
From the above diagram, $\sin θ = \frac{h}{x}$
$\begin{aligned} x = \frac{h}{\sin θ} = \frac{1.2}{\sin 30^{\circ}} \\ = 1.2 \times 2 = 2.4 m \end{aligned}$

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