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Chapter 1: Q44E (page 297)

A bucket of mass m is tied to a massless cable that is wrapped around the outer rim of a frictionless uniform pulley of radius R , similar to the system shown in Fig.49.43. In terms of the stated variables what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket?

Short Answer

Expert verified

Thus, Kinetic energy as the bucket is\(\frac{1}{2}m{R^2}\).

Step by step solution

01

Step:-1 explanation

Given in the question A bucket of mass m is tied to a massless cable that is wrapped around the outer rim of a frictionless uniform pulley of radius R.

02

Step:-2 Concept

We know that the formula is used for finding different kinetic energy for the bucket,\({K_p} = \frac{1}{2}I{\omega ^2}\)and\({K_b} = \frac{1}{2}m{v^2}\).

\({K_p} = \frac{1}{2}{K_b}\).

03

Step:-3 calculation

The speed of the bucket and the pulley are related\(v = R\omega \).

We know that\({K_p} = \frac{1}{2}{k_b}\)___________(1)

Put the value\({K_p}\)and\({K_b}\)in equation (1)

We get ,

\(\frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{1}{2}m{v^2}} \right)\)

\(\frac{1}{2}I{\omega ^2} = \frac{1}{4}m{v^2}\)

\(I = \frac{1}{2}m{R^2}\)

Hence, kinetic energy is the bucket \(\frac{1}{2}m{R^2}\)

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