Question

Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform angular acceleration of 0.01 rad/s2 about the center O. In the figure, the magnitude of the linear acceleration of P, when it reaches the y-axis, is closest to:
a..063
b..075
c..072
d..069
e..066

Answer 1

Answer:

e). $a = 0.066 m/s^2$

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

$\omega_f^2 - \omega_i^2 = 2\alpha \theta$

now we have

$\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})$

$\omega = 0.177 rad/s$

now the centripetal acceleration of the point P is given as

$a_c = \omega^2 R$

$a_c = (0.177)^2(2)$

$a_c = 0.063 m/s^2$

tangential acceleration is given as

$a_t = R\alpha$

$a_t = 2(0.01)$

$a_t = 0.02 m/s^2$

now net acceleration is given as

$a = \sqrt{a_t^2 + a_c^2}$

$a = \sqrt{0.02^2 + 0.063^2}$

$a = 0.066 m/s^2$