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Q88P

Expert-verifiedFound in: Page 293

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A thin spherical shell has a radius of 1.90 m.**** An applied torque of 960 N.m** **gives the shell an angular acceleration of 6.20 rad/s ^{2}**

** **

- Rotational inertia of the shell about axis is $154.8\u200a\u200a\text{kg}.{\text{m}}^{2}.$
- Mass of the shell is data-custom-editor="chemistry" $64.3\text{kg.}$

- Radius of the thin shell is $1.90m$
- Torque is $960\text{N}$
- Angular acceleration is $6.20\text{rad}/{\text{s}}^{2}$

**Use the basic formula for torque in terms of inertia and angular acceleration to find the rotational inertia. Mass can be found from the value of rotational inertia using the formula for the rotational inertia in terms of mass and radius.**

Formulae are as follow:

$\tau =I\times \alpha $

$I=\frac{2}{3}M{R}^{2}$

Where,

$\tau $ is torque, *M* is mass, *R* is radius, *I* is moment of inertia and $\alpha $ is angular acceleration.

By using formula for torque as follows,

$\tau =I\times \alpha \phantom{\rule{0ex}{0ex}}960=I\times 6.20\phantom{\rule{0ex}{0ex}}I=154.8\text{kg}-{\text{m}}^{2}$

Hence, rotational inertia of the shell about axis is $154.8\text{kg}-{\text{m}}^{2}$

Now, using the following formula, mass can be found,

$I=\frac{2}{3}M{R}^{2}$

This is for a spherical shell,

$154.8=\frac{2}{3}M\times 1.{90}^{2}$

$M=64.3\text{kg}$

Hence, mass of the shell is $64.3\text{kg.}$

Therefore, from the given torque and angular acceleration, the rotational inertia can be found. Using the formula for rotational inertia of the sphere, the mass of the sphere can be found.

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