Question

A grinding wheel is a uniform cylinder with a radius of 8.20 cm and a mass of 0.580 kg.
(a) Calculate its moment of inertia about its center.
___kg·m2
(b) Calculate the applied torque needed to accelerate it from rest to 1200 rpm in 5.00 s if it is known to slow down from 1200 rpm to rest in 56.0 s.
___m·N

Answer 1

Concepts and reason

The concepts used to solve the problem are torque and moment of inertia of a cylinder.

First, obtain the expression for moment of inertia of a cylinder relating its mass and radius. Later, determine angular acceleration from values of angular velocities and time provided.

Finally, calculate torque acting to the cylinder by using values of the moment of inertia and angular acceleration.

Fundamentals

The quantity which resists the body’s angular acceleration is called its moment of inertia.

Moment of inertia of a cylinder about its ends is calculated by below expression

$I = \frac{1}{2}m{r^2}$

Here, the moment of inertia is $I$ , the mass of the cylinder is $m$ and the radius of gyration is $r$ .

When a torque is applied to an object it begins to rotate.

A relation between a torque, moment of inertia and angular acceleration is

$\tau \, = I\alpha$

Here, the torque on the rotating object is $\tau \,$ , the moment of inertia is $I$ and angular acceleration of the rerating object is $\alpha$ .

Angular acceleration is calculated as follows,

$\alpha \, = \,\frac{{{\omega _{\rm{f}}}\, - {\omega _{\rm{i}}}\,}}{{\Delta t}}$

Here, the initial angular velocity is ${\omega _i}\,$ , the final angular velocity is ${\omega _f}\,$ , and time elapsed between two angular velocities is $\Delta t$ .

(a)

A grinding wheel can be considered as a cylinder. Hence its moment of inertia has the expression as follows,

$I = \frac{1}{2}m{r^2}$

Here, the mass of the cylinder is $m$ and the radius of the cylinder is $r$ .

Substitute $0.58\,\,{\rm{kg}}$ for $m$ and $8.20\,{\rm{cm}}$ .

$\begin{array}{l}\\I = \left( {0.58\,{\rm{kg}}} \right){\left( {\left( {8.20\,{\rm{cm}}} \right)\left( {\frac{{1\,\,{\rm{m}}}}{{{{10}^2}\,{\rm{cm}}}}} \right)} \right)^2}\\\\I = 0.00195\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\\\end{array}$

(b)

Angular acceleration in terms of initial and final angular velocities and time is given as below,

$\alpha \, = \,\frac{{{\omega _f}\, - {\omega _i}\,}}{{\Delta t}}$

Here, the initial angular velocity is ${\omega _i}\,$ , the final angular velocity is ${\omega _f}\,$ , and time elapsed between two angular velocities is $\Delta t$ .

Substitute $0\,{\rm{rpm}}$ for ${\omega _i}\,$ , $1200\,{\rm{rpm}}$ for ${\omega _f}\,$ and $5.00\,{\rm{s}}$ for $\Delta t$ .

$\begin{array}{l}\\\alpha \, = \,\left( {\frac{{1200\,\frac{{{\rm{rev}}\,}}{{\min }} - 0\,\frac{{{\rm{rev}}\,}}{{\min }}}}{{5.00\,{\rm{s}}}}} \right)\left( {\frac{{2\pi \,{\rm{rad}}}}{{1\,{\rm{rev}}}}} \right)\left( {\frac{{1\,\min }}{{60\,s}}} \right)\\\\\alpha \, = 25.14\,\,\frac{{{\rm{rad}}\,}}{s}\\\end{array}$

The torque is expressed in terms of the moment of inertia and angular acceleration as below,

$\tau \, = I\alpha$

Here, the torque is $\tau \,$ , the moment of inertia is $I$ and the angular acceleration is $\alpha$ .

Substitute $0.00195\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}$ for $I$ and $25.14\,\,\frac{{{\rm{rad}}\,}}{{{s^2}}}$ for $\alpha$ .

$\begin{array}{l}\\\tau \, = \left( {0.00195\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {25.14\,\,\frac{{{\rm{rad}}\,}}{{{s^2}}}} \right)\\\\\tau \, = 0.049\,\,\frac{{{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {\rm{rad}}}}{{{s^2}}}\\\\\tau \, = 0.049\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

Ans: Part a

The moment of inertia of the grinding wheel is ${\bf{0}}{\bf{.00195}}\,{\bf{kg}} \cdot {{\bf{m}}^{\bf{2}}}$ .