Question

A flywheel with a radius of 0.700 m starts from rest and accelerates with a constant angular acceleration of 0.800 rad/s2 . part a part complete compute the magnitude of the tangential acceleration of a point on its rim at the start. atan = 0.560 m/s2 submitprevious answers correct part b part complete compute the magnitude of the radial acceleration of a point on its rim at the start. arad = 0 m/s2 submitprevious answers correct part c part complete compute the magnitude of the resultant acceleration of a point on its rim at the start. a = 0.560 m/s2 submitprevious answers correct part d compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 60.0 â.

Answer 1

1) For a point on the rim of the flywheel, the distance from the center of the motion is equal to the radius of the wheel: $r=0.700 m$.

The tangential acceleration for a point on the rim is given by
$a_t = \alpha r$
where $\alpha=0.800 rad/s^2$ is the angular acceleration while r is the radius. Substituting the numbers, we get
$a_t = (0.800 rad/s^2)(0.700 m)=0.560 m/s^2$

2) The radial acceleration for a point on the rim (r=0.700 m) at time t=0 is given by:
$a_r = \frac{v_t^2}{r}$
where $v_t$ is the tangential velocity at time t=0.

The tangential velocity is given by
$v_t = \omega r$
where $\omega$ is the angular speed; however, at the start of the motion (t=0) the flywheel is at rest, so $\omega=0$ and $v_t=0$, so the radial acceleration is $a_r = 0$.

3) The magnitude of the acceleration at time t=0 is given by:
$|a| = \sqrt{a_t^2 + a_r^2}= \sqrt{(0.560 m/s^2)^2+(0)^2} =0.560 m/s^2$

4) As we said at point 1), the tangential acceleration is given by
$a_t = \alpha r$
but $\alpha$, the angular acceleration, is constant, so the tangential acceleration after an angle of $\theta=60.0^{\circ}$ is just equal to the tangential acceleration at the beginning of the motion:
$a_t = (0.800 rad/s^2)(0.700 m)=0.560 m/s^2$