Question

A 1.6-kg grindstone in the shape of a uniform cylinder of radius 0.20 m acquires a rotational rate of \({\bf{24}}\;{{{\bf{rev}}} \mathord{\left/

{\vphantom {{{\bf{rev}}} {\bf{s}}}} \right.

\\{\bf{s}}}\)from rest over a 6.0-s interval at constant angular acceleration. Calculate the torque delivered by the motor.

Short answer

The torque developed by the motor is \(0.80\;{\rm{N}} \cdot {\rm{m}}\).

Step-by-step solution

Determination of torque

Torque may be defined as the product of the object’s angular acceleration and moment of inertia. Its value is altered linearly to the value of the angular acceleration of the object.

Given information

Given data:

The mass of the grindstone is\(m = 1.6\;{\rm{kg}}\).

The radius of the grindstone is\(r = 0.20\;{\rm{m}}\).

The final angular velocity is\({\omega _2} = 24\;{{{\rm{rev}}} \mathord{\left/

{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.

\\{\rm{s}}}\).

Since the grindstone starts from rest, the initial angular velocity is\({\omega _1} = 0\).

The time interval is \(\Delta t = 6.0\;{\rm{s}}\).

Evaluation of moment of inertia of the grindstone

The moment of inertia of a solid cylinder or grindstone can be calculated as:

\(\begin{aligned}{c}I = \frac{1}{2}m{r^2}\\I = \frac{1}{2}\left( {1.6\;{\rm{kg}}} \right){\left( {0.20\;{\rm{m}}} \right)^2}\\I = 0.032\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{aligned}\)

Evaluation of the angular acceleration of the grindstone

The angular acceleration of the grindstone can be calculated as:

\(\begin{aligned}{l}\alpha = \frac{{{\omega _2} - {\omega _1}}}{{\Delta t}}\\\alpha = \frac{{\left[ {\left( {24\;{{{\rm{rev}}} \mathord{\left/

{\vphantom {{{\rm{rev}}} {\rm{s}}}} \right.

\\{\rm{s}}}} \right)\left( {\frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}}} \right)} \right] - 0}}{{\left( {6.0\;{\rm{s}}} \right)}}\\\alpha = 25.13\;{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {{{\rm{s}}^2}}}} \right.

\\{{{\rm{s}}^2}}}\end{aligned}\)

Evaluation of the torque delivered by the motor

The torque developed by the motor can be calculated as:

\(\begin{aligned}{c}\tau = I\alpha \\\tau = \left( {0.032\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {25.13\;{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {{{\rm{s}}^2}}}} \right.

\\{{{\rm{s}}^2}}}} \right)\\\tau = 0.80\;{\rm{N}} \cdot {\rm{m}}\end{aligned}\)

Thus, the torque developed by the motor is \(0.80\;{\rm{N}} \cdot {\rm{m}}\).