Question

A machinist turns on the power on to a grinding wheel at time t= 0 s. The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 58rad/s. The wheel is run at that angular velocity for 30 s, and then power is shut off. The wheel slows down uniformly at 1.4rad/s^2 until the wheel stops. What is the total number of revolutions made by the wheel in this situation?A) 340 B) 390 C) 430 D) 570 E) 300

Answer 1

The total number of revolutions made by the wheel is 514

Explanation:

We can solve the problem by applying the suvat equations for rotational motion, to the two different parts of the motion.

During the first part (acceleration), we have:

$\omega_0 =0$ (initial angular velocity)

$\omega_1=58 rad/s$ (final angular velocity)

$t_1=10 s$ (time)

So the angular displacement covered in this part is

$\theta_1 = (\frac{\omega_0+\omega_1}{2})t_1 =(\frac{0+58}{2})(10)=290 rad$

In the second part, we have uniform (circular) motion, with constant angular velocity

$\omega_2 = 58 rad/s$

for

t = 30 s

So the angular displacement in this part is

$\theta_2 = \omega_2 t_2 = (58)(30)=1740 rad$

In the third part, we have a decelerated motion, with constant angular acceleration of

$\alpha=-1.4 rad/s^2$

and initial angular velocity

$\omega_2 = 58 rad/s$

while final angular velocity is

$\omega_3 = 0$

So the angular displacement in this part is given by

$\omega_3^2 - \omega_2^2 = 2\alpha \theta_3\\\theta_3 = \frac{\omega_3^3-\omega_2^2}{2\alpha}=\frac{0-58^2}{2(-1.4)}=1201 rad$

So the total angular  displacement of the wheel is

$\theta=290 + 1740 + 1201 = 3231 rad$

Converting into revolutions,

$\theta=3231 rad \cdot \frac{1}{2\pi rad/rev}=514 rev$

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