Rotational motion

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Question

Two beads each of mass m are welded at the ends of two light rigid rods each of length l. If the pivots are smooth then find the ratio of translational and rotational kinetic energy of system.

Moderate

Solution

We know translational kinetic energy $K_{t} = \frac{1}{2} {mv}_{C}^{2}$ and rotational kinetic energy $K_{r} = \frac{1}{2} I_{C} ω^{2}$
Distance of centre of mass from pivot end,
$r_{C} = \frac{ml + m (2 l)}{m + m} = \frac{3}{2} l$
Moment of inertia of system about centre of mass,
$I_{C} = m {(\frac{l}{2})}^{2} + m {(\frac{l}{2})}^{2} = \frac{{ml}^{2}}{2}$
Angular velocity of rod $ω = \frac{v}{2 l}$
velocity of centre of mass, $v_{cm} = ω (\frac{3}{2} l) = (\frac{v}{2 l}) (\frac{3}{2} l) = \frac{3}{4} v$
Translational kinetic energy of system,
$K_{transiational} = \frac{1}{2} M_{total} \cdot v_{cm}^{2} = \frac{1}{2} (2 m) {(\frac{3}{4} v)}^{2} = \frac{9}{16} {mv}^{2}$
$K_{roltaional} = \frac{1}{2} I_{CM} ω^{2} = \frac{1}{2} (\frac{{ml}^{2}}{2}) {(\frac{v}{2 l})}^{2} = \frac{1}{16} {mv}^{2}$
This gives $\frac{K_{t}}{K_{r}} = 9$ .

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Similar Questions

The condition of equilibrium is satisfied when resultant of all forces acting on body vanishes and also no net torque acts on the body. However body can be different types of equilibrium as stable, unstable, neutral or may be having pure rotation when no resultant force acts on the body. Different conditions of equilibrium or rotational motion can be known by different types of energy it possesses. Column I mentions the state of body and column II mentions the type of energy it possesses. Match the correct options in two columns

COLUMN_I COLUMN_II

A) Body in stable equilibrium P) Rotational kinetic energy

B) Body in unstable equilibrium Q) Minimum potential

energy

C) Body in neutral equilibrium R)Maximum potential energy

D) Body in pure rotational motion S) Constant potential energy

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