A CD with a diameter of 12.0 cm starts from rest and with a constant angular acceleration of 1.0 rad/sec² acquires an angular velocity of 5.0 rad/sec.The CD continues rotating at 5.0 rad/sec for 15.0 seconds and then slows to a stop in 12.0 second with a constant angular deceleration.What is the centripetal acceleration of a point 4.0 cm from the center at the time 15.0 seconds from the start? A) 0.5 m/s² B) 1.0 m/s² C) 1.8 m/s² D) 2.0 m/s² E) 2.6 m/s²

Question

Quizplus · Accepted Answer

First let's find the final angular velocity after the initial acceleration:

$\omega_f = \omega_i + \alpha t$
$\omega_f = 0 + (1.0\,	ext{rad/s}^2)(15\,	ext{s}) = 15\,	ext{rad/s}$

Next, let's find the angular displacement during the first part of the motion:

$	heta = \dfrac{1}{2}\alpha t^2$
$	heta = \dfrac{1}{2}(1.0\,	ext{rad/s}^2)(15\,	ext{s})^2 = 112.5\,	ext{rad}$

The angular displacement during the second part of the motion is just the remaining portion of a full rotation:

$	heta = 2\pi - 112.5\,	ext{rad} = 1.57\,	ext{rad}$

Now we can find the time and angular acceleration during the deceleration:

$t = 12\,	ext{s}$

$\alpha = -\dfrac{\omega_f}{t} = -\dfrac{15\,	ext{rad/s}}{12\,	ext{s}} = -1.25\,	ext{rad/s}^2$

Finally, we can use the centripetal acceleration formula:

$a_c = r\omega^2$

At 15.0 seconds:

$a_c = (0.04\,	ext{m})(15.0\,	ext{rad/s})^2 = 9\,	ext{m/s}^2$

At the end of the motion:

$a_c = (0.04\,	ext{m})(5.0\,	ext{rad/s})^2 = 1\,	ext{m/s}^2$

The average acceleration during the deceleration is:

$a_{	ext{avg}} = \dfrac{\Delta\omega}{\Delta t} = \dfrac{5\,	ext{rad/s}}{12\,	ext{s}} = 0.42\,	ext{rad/s}^2$

Using the equation $v^2 = v_0^2 + 2a\Delta x$, we can find the distance traveled during the deceleration:

$\Delta x = \dfrac{v_f^2 - v_0^2}{2a} = \dfrac{-25\,	ext{rad/s}^2}{2(-1.25\,	ext{rad/s}^2)} = 10\,	ext{s}$

Thus, the total time for the motion is:

$15\,	ext{s} + 12\,	ext{s} + 10\,	ext{s} = 37\,	ext{s}$

And the total angular displacement is:

$112.5\,	ext{rad} + 1.57\,	ext{rad} + (15\,	ext{rad/s})(10\,	ext{s}) + \dfrac{1}{2}(-1.25\,	ext{rad/s}^2)(10\,	ext{s})^2 = 474\,	ext{rad}$

So the centripetal acceleration at a point 4.0 cm from the center at 15.0 seconds is:

$a_c = r\omega^2 = (0.04\,	ext{m})(15.0\,	ext{rad/s})^2 = 9\,	ext{m/s}^2 \approx \boxed{	extbf{(B) }1.0\,	ext{m/s}^2}$.

A CD with a Diameter of 12