Question

A 250 , 25-cm-diameter plastic disk is spun on anaxle through its center by an electric motor.

What torque must the motor supply to take the disk from 0to 1600 rpm in 4.8 ?

Express your answer using two significant figures.(answer?=N*m)

Answer 1

Concepts and reason

The concepts used to solve this problem are angular kinematics and moment of inertia.

Initially, the angular acceleration can be calculated by using the angular kinematics formula. Then, the moment of inertia can be calculated for the disk by using its formula. Finally, the torque acting on the system can be calculated by multiplying the moment of inertia and angular acceleration.

Fundamentals

The expression for the angular acceleration from angular kinematics is,

${\omega _f} = {\omega _i} + \alpha t$

Here, ${\omega _f}$ is the final angular speed, ${\omega _i}$ is the initial angular speed, $\alpha$ is the angular acceleration, and $t$ is the time period.

The moment of inertia for the disk is,

$I = \frac{1}{2}M{R^2}$

Here, $I$ is the moment of inertia, $M$ is the mass of the disk, and $R$ is the radius of the disk.

The formula for the torque is,

$\tau = I\alpha$

Here, $\tau$ is the torque acting on the system.

Rpm is revolution per min which is expressed as ${\rm{rpm}} = \frac{{{\rm{rev}}}}{{\min }}$ .

The expression for the angular acceleration is,

${\omega _f} = {\omega _i} + \alpha t$

Substitute $1600\;{\rm{rpm}}$ for ${\omega _f}$ , $0$ for ${\omega _i}$ , 4.8s for $t$ .

$\begin{array}{c}\\0 + 4.8\alpha = \left( {1600\,\frac{{{\rm{rev}}}}{{\min }}} \right)\left( {\frac{{1\min }}{{60\,{\rm{s}}}}} \right)\left( {2\pi \,{\rm{rad}}} \right)\\\\167.47 = 4.8\alpha \\\\\alpha = 34.9\,{\rm{rad/}}{{\rm{s}}^2}\\\end{array}$

The radius of the plastics disk is,

$R = \frac{d}{2}$

Here, d is the diameter of the disk.

Substitute $25\,{\rm{cm}}$ for d to find R.

$\begin{array}{c}\\R = \frac{{\left( {25\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)}}{2}\\\\ = 0.125\,{\rm{m}}\\\end{array}$

The expression for the moment of inertia is,

$I = \frac{1}{2}M{R^2}$

Substitute $250{\rm{g}}$ for $M$ and $0.125\,{\rm{m}}$ for $R$ .

$\begin{array}{c}\\I = \frac{1}{2}\left( {250\,{\rm{g}}} \right)\left( {\frac{{1 \times {{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right){\left[ {\left( {12.5\,{\rm{cm}}} \right)\left( {\frac{{1 \times {{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)} \right]^2}\\\\ = 1.95 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\\\end{array}$

The expression for the torque is,

$\tau = I\alpha$

Substitute $1.95 \times {10^{ - 3}}\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}$ for $I$ and $34.9\,{\rm{ rad/}}{{\rm{s}}^2}$ for $\alpha$ .

$\begin{array}{c}\\\tau = \left( {1.95 \times {{10}^{ - 3}}\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( {34.9\,{\rm{rad/}}{{\rm{s}}^2}} \right)\\\\ = 0.0681\,{\rm{N}} \cdot {\rm{m}}\\\\ \approx {\rm{0}}{\rm{.068}}\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

Ans:

The torque acting on the system is $0.068\;{\rm{N}} \cdot {\rm{m}}$ .