Question

TA phonograph turntable rotating at $78rev/\min $ slows down and stops in 30 sec after the motor is turned off. Then the revolutions in $rev/\min $ made by it in this time are:
A. 19.5
B. 39
C. 78
D. 156

Answer 1

Hint: Here we have to write the revolutions $\theta $ by assuming the given initial angular velocity and find the time after 30 sec. here we have to find $\alpha $ which is angular acceleration. And as the phonograph turntable speed slows down such that angular acceleration should be negative.

Complete solution step by step:
As given above that initial angular velocity ${{\omega }_{i}}=78rpm$ and also the final angular velocity will be,
${{\omega }_{f}}=0rpm$

As the phonograph speed slows down hence after 30 sec it stops thus the given time will be $0.5\min $ .
Or,
$t=\dfrac{30}{60}\min $
$\Rightarrow t=0.5\min $

Now the angular acceleration is given as,
$\alpha =\dfrac{{{\omega }_{f}}-{{\omega }_{i}}}{t}$
$\alpha =\dfrac{0-78}{0.5}$
$\alpha =-156rev/{{\min }^{2}}$

It is negative because the speed of the phonograph slows down and angular velocity of the turn-table decreases over time.
Now,
$\theta ={{\omega }_{i}}t+\dfrac{1}{2}\alpha {{t}^{2}}$ eq. (1)

Now on putting the value of initial angular velocity, angular acceleration and time. We find revolutions in $rev/\min $ made by phonograph in 30 sec is,
$\theta =78\times 0.5+\dfrac{1}{2}(-156){{(0.5)}^{2}}$
$\theta =39-19.5$
$\theta =19.5rev$

Hence the revolutions in time 30 sec is $19.5rev$. That is option A.

Note:
Phonograph is made in $1900$ to $1925$ and it is observed that it revolves $74$ and $82$ revolutions per minute, then they stabilize at 78 revolutions per minute with the electrical table. The synchronous motor ran at 3600 revolution per minute. With the ratio of $46:1$ gear, this produces a speed of $78.26$ revolution per minute, which became the standard value for the revolution of phonograph.