Question

Review. A piece of putty is initially located at pointon the rim of a grinding wheel rotating at constant angular speed about a horizontal axis. The putty is dislodged from point Awhen the diameter through Ais horizontal. It then rises vertically and returns to $\mathit{A}$at the instant the wheel completes one revolution. From this information, we wish to find the speed$\mathit{v}$of the putty when it leaves the wheel and the force holding it to the wheel. (a) What analysis model is appropriate for the motion of the putty as it rises and falls? (b) Use this model to find a symbolic expression for the time interval between when the putty leaves point $\mathit{A}$and when it arrives back at$\mathit{A}$, in terms of$\mathit{v}$and$\mathit{g}$. (c) What is the appropriate analysis model to describe point$\mathit{A}$on the wheel? (d) Find the period of the motion of point$\mathit{A}$in terms of the tangential speed$\mathit{v}$and the radius$\mathit{R}$of the wheel. (e) Set the time interval from part (b) equal to the period from part (d) and solve for the speedof the putty as it leaves the wheel. (f) If the mass of the putty is$\mathit{m}$, what is the magnitude of the force that held it to the wheel before it was released?

Short answer

(a) This analysis model is appropriate by considering the point A in projectile motion when the putty as it rises and falls.

(b) The expression for the time t in terms of v and g is$t=\frac{2v}{g}$

(c) The analysis model to describe the motion of the point A on the wheel is uniform circular motion.

(d) The period of the motion of point A in terms of tangential speed v and the radius R of the wheel is role="math" localid="1663683935002" $t=\frac{2\pi R}{v}$.

(e) The speed of the putty as it leaves the wheel is$v=\sqrt{g\pi R}$

(f) The magnitude of the force that held it to the wheel before it was released is $F=\pi \left(mg\right)$.

Step-by-step solution

Centrifugal force.

The expression for the centrifugal force is given by,

$F=m\frac{v^2}{r}$

Here is the mass of the object, v is the speed and r is the radius, F is the centrifugal force. Centrifugal force is also known by the name of centripetal force.

The third law of the motion is given by,

$v_f=v_i+at$

Here$V_f$ is the final velocity,$V_i$ is the final velocity, is the acceleration and t is the time.

State the analysis model which is appropriate for the motion of the putty as it rises and falls.

(a)

We can assume the piece of putty as a particle which revolves around a horizontal axis in circular motion

Point A is in projectile motion is appropriate model for the motion of the putty as it rises and falls.

Find the expression for the time

(b)

The third law of motion is given by,

$v_f=v_i+a_{y}t$

As, the final velocity is negative of initial velocity.

$$\begin{gathered} -v=v=v-gt \\ gt=2v \end{gathered}$$

$\therefore t=\frac{2v}{g}$....... (1)

Therefore the expression for the time in terms of v and g is$t=\frac{2v}{g}$.

Explanation for (c).

(c)

The analysis model to describe the motion of the point on the wheel is uniform circular motion.

Calculate the time period of the motion of point A.

(d)

The expression for the speed in terms of angular velocity and radius is given by,

$v=\omega r$

Here v is the speed, $\omega$is the angular speed and r is the radius.

Substitute $\frac{2\pi }{T}$for $\omega$in the above equation.

$v=\left(\frac{2\pi }{T}\right)r$

$\therefore T=\frac{2\pi r}{v}$...... (2)

Therefore the period of the motion of point A in terms of tangential speed v and the radius R of the wheel is$T=\frac{2\pi R}{v}$.

Calculate the speed of putty.

(e)

Equating equation (1) and (2)

$$\begin{gathered} \frac{2v}{g}=\frac{2\pi R}{v} \\ {v}^2=g\pi R \\ v=\sqrt{g\pi R} \end{gathered}$$

Hence, the speed of the putty as it leaves the wheel is$v=\sqrt{g\pi R}$.

Calculate the magnitude of force that held it to the wheel before it was released.

(f)

If the mass of putty is then the radial force that held it to the wheel is

$F=\frac{m{v}^2}{R}$

Substitute $\sqrt{g\pi R}$in the above equation.

$$\begin{gathered} F=\frac{m\left(g\pi R\right)}{R} \\ F=\pi mg \end{gathered}$$

Hence, the magnitude of the force that held it to the wheel before it was released is$F=\pi mg$ .