A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 and has a constant angular acceleration of 2.1 rad/s². At what value of t will the radial and tangential components of the linear acceleration of a point on the rim of the disk be equal in magnitude? A) 0.55 s B) 0.63 s C) 0.69 s D) 0.59 s E) 0.47 s

Question

Quizplus · Accepted Answer

The radial component of the linear acceleration is given by $a_r = r\alpha$ where $r$ is the radius and $\alpha$ is the angular acceleration. The tangential component of the linear acceleration is given by $a_t = r\omega^2$ where $\omega$ is the angular velocity.
At the instant when the magnitudes of the radial and tangential components are equal, we have $a_r = a_t$. Substituting the above expressions, we get $r\alpha = r\omega^2$, which simplifies to $\alpha = \omega^2$.
The angular velocity as a function of time is given by $\omega = \omega_0 + \alpha t$, where $\omega_0$ is the initial angular velocity (which is zero, since the disk starts from rest). Substituting the given value of the angular acceleration, we get $\omega = 2.1t$.
Now we can set up the equality $\alpha = \omega^2$ and solve for $t$: 
$$2.1 = (2.1t)^2$$
$$2.1 = 4.41t^2$$
$$t^2 = \frac{2.1}{4.41}$$
$$t \approx 0.695$$
Therefore, the answer is choice C, 0.69 s (rounded to two significant figures).

A Horizontal Disk with a Radius of 10 Cm Rotates