91Ó°ÊÓ

Torque is the product of the moment of inertia and the angular acceleration. For this problem, the net torque equals the sum of the applied torque and the frictional torque.

The total mass of the cylindrical wheel is \(m = 0.380\;{\rm{kg}}\).

The radius of the cylindrical wheel is \(r = 8.50\;{\rm{cm}} = 0.085\;{\rm{m}}\).

The initial angular velocity of the wheel is \({\omega _1} = 0\).

The final angular velocity of the wheel is \({\omega _2} = 1750\;{\rm{rpm}} = \left( {\frac{{1750 \times 2\pi }}{{60}}} \right)\;{\rm{rad/s}}\).

The time for the rotation of the wheel is \(t = 5.00\;s\).

Now, the frictional torque takes \(t' = 55.0\;{\rm{s}}\) to come to rest from the initial angular velocity of\({\omega _3} = 1500\;{\rm{rpm}} = \left( {\frac{{1500 \times 2\pi }}{{60}}} \right)\;{\rm{rad/s}}\).

Let \(\tau \) be the applied torque on the wheel.

Part (a)

The moment of inertia of the cylindrical wheel is

\(\begin{align}I &= \frac{1}{2}m{r^2}\\ &= \frac{1}{2} \times \left( {0.380\;{\rm{kg}}} \right) \times {\left( {0.085\;{\rm{m}}} \right)^2}\\ &= 1.37 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{align}\).

Hence, the moment of inertia of the wheel is \(1.37 \times {10^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

Part (b)

The frictional torque on the wheel is \({\tau _{\rm{f}}} = I{\alpha _{\rm{f}}}\).

Here, \({\alpha _{\rm{f}}}\) is the angular acceleration due to the frictional torque, and the value of \({\alpha _{\rm{f}}}\) is

\(\begin{align}{\alpha _{\rm{f}}} &= \frac{{0 - {\omega _3}}}{{t'}}\\ &= \frac{{ - {\omega _3}}}{{t'}}\end{align}\).

Therefore, the frictional torque on the wheel is

\(\begin{align}{\tau _{\rm{f}}} &= - I\frac{{{\omega _3}}}{{t'}}\\ &= - \left( {1.37 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) \times \frac{{\left( {\frac{{1500 \times 2\pi }}{{60}}} \right)\;{\rm{rad/s}}}}{{55.0\;{\rm{s}}}}\\ &= - 3.91 \times {10^{ - 3}}\;{\rm{N}} \cdot {\rm{m}}\end{align}\).

The net angular acceleration of the wheel is \({\alpha _{{\rm{net}}}} = \frac{{{\omega _2} - {\omega _1}}}{t}\).

Now, the net torque on the wheel is

\(\begin{align}{\tau _{{\rm{net}}}} &= I{\alpha _{{\rm{net}}}}\\ &= \left( {1.37 \times {{10}^{ - 3}}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right) \times \frac{{\left( {\frac{{1750 \times 2\pi }}{{60}}} \right)\;{\rm{rad/s}} - 0}}{{5.0\;{\rm{s}}}}\\ &= 50.19 \times {10^{ - 3}}\;{\rm{N}} \cdot {\rm{m}}\end{align}\)

Now,

\(\begin{align}{\tau _{{\rm{net}}}} &= \tau + {\tau _{\rm{f}}}\\\tau &= {\tau _{{\rm{net}}}} - {\tau _{\rm{f}}}\\ &= \left( {50.19 \times {{10}^{ - 3}}\;{\rm{N}} \cdot {\rm{m}}} \right) - \left( { - 3.91 \times {{10}^{ - 3}}\;{\rm{N}} \cdot {\rm{m}}} \right)\\ &= 5.41 \times {10^{ - 2}}\;{\rm{N}} \cdot {\rm{m}}\end{align}\).

Hence, the applied torque on the wheel is \(5.41 \times {10^{ - 2}}\;{\rm{N}} \cdot {\rm{m}}\).