Q94PP 94. The moment of inertia of the... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 1: Q94PP (page 337)

94. The moment of inertia of the empty turntable is $1.5 {kgm}^{2}$ . With a constant torque of $2.5 N 鈰� m$ , the turntable鈥損erson system takes $3.0 s$ to spin from rest to an angular speed of $1.0 rad / s$ . What is the person鈥檚 moment of inertia about an axis through her center of mass? Ignore friction in the turntable axlerole="math" localid="1667995844298" $(a) 2.5 {kgm}^{2}; (b) 6.0 {kgm}^{2}; (c) 7.5 {kgm}^{2}; (d) 9.0 {kgm}^{2}$ .

Short Answer

Expert verified

The angular momentum of the person is $I_{_{p e r s o n}} = 6.0 {kgm}^{2}$ . Hence, option b is correct.

Step by step solution

Given Data

It is given that the moment of inertia of empty turntable as $I_{0} = 1.5 {kgm}^{2}$ , constant torque as $蟿 = 2.5 N 鈰� m$ , time as $t = 3.0 s$ and final angular speed as $蝇 = 1.0 r a d / s$ .

Angular acceleration

The angular acceleration is produced by the constant torque as $蟿 = I 伪$ 鈥︹€� (1)

Find $a$ from the equation $蝇 = 蝇_{0} + 伪 t$ . Since, the initial angular speed is zero that is $蝇_{0} = 0 then 伪 = \frac{蝇}{蟿}$ .

Substitute $伪$ in (1) then, $蟿 = \frac{I 蝇}{蟿}$ .

Angular momentum of the system

The total angular momentum of the system is $I = I_{0} + I_{p e r s o n}$ . Then $蟿$ can be written as

$蟿 = (I_{0} + I_{p e r s o n}) \frac{蝇}{t} 鈬� I_{0} + I_{p e r s o n} = \frac{r t}{蝇} 鈬� I_{p e r s o n} = \frac{r t}{蝇} = I_{0}$

Substitute all the known values and simplify.

$I_{p e r s o n} = \frac{蟿 t}{蝇} - I_{0} = \frac{(2.5 Nm) (3.0 s)}{1.0 rad / s} - 1.5 {kgm}^{2} = 7.5 {kgm}^{2} - 1.5 {kgm}^{2} = 6.0 {kgm}^{2}$

Therefore, the angular momentum of the person is $I_{p e r s o n} = 6.0 {kgm}^{2}$ .

Hence, option b is correct.

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Short Answer

Step by step solution

Given Data

Angular acceleration

Angular momentum of the system

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