Rotational kinematics with constant angular acceleration: In the figure, point P is at rest when it is on the x-axis. The linear speed of point P when it reaches the y-axis is closest to
A) 0.18 m/s.
B) 0.24 m/s.
C) 0.35 m/s.
D) 0.49 m/s.
E) 0.71 m/s.

Question

Rotational kinematics with constant angular acceleration: In the figure, point P is at rest when it is on the x-axis. The linear speed of point P when it reaches the y-axis is closest to 
A) 0.18 m/s. 
B) 0.24 m/s. 
C) 0.35 m/s. 
D) 0.49 m/s. 
E) 0.71 m/s.

Quizplus · Accepted Answer

Using the rotational kinematics equations, we have:

$	heta = 	heta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$

$\omega = \omega_0 + \alpha t$

$v = r\omega$

Assuming counterclockwise rotation to be positive, we can use the given information to get:

$	heta_0 = 0$

$	heta = \frac{\pi}{2}$

$\omega_0 = 0$

$v = ?$

Now we need to find $\alpha$ in order to be able to solve for $t$ using $	heta$:

$	heta = 	heta_0 + \frac{1}{2}\alpha t^2$

$\frac{\pi}{2} = \frac{1}{2}\alpha t^2$

$t^2 = \frac{\pi}{\alpha}$

Next, we can find $\omega$ using:

$\omega = \omega_0 + \alpha t$

$\omega = \alpha t$

$\omega = \sqrt{\alpha\cdot\frac{\pi}{\alpha}} = \sqrt{\pi}$

Finally, we can find $v$ using:

$v = r\omega$

$v = 0.25\cdot\sqrt{\pi}\approx 0.35	ext{ m/s}$

Therefore, the answer is (C).

Rotational Kinematics with Constant Angular Acceleration: in the Figure, Point