Question

At a time t = 3.10 s , a point on the rim of a wheel with a radius of 0.200 m has a tangential speed of 50.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.3 m/s2 .

Answer 1

Answer:

Angular acceleration, $\alpha =51.5\ rad/s^2$

Explanation:

It is given that,

Radius of the wheel, r = 0.2 m

Tangential speed of the wheel, $v=50\ m/s$

Tangential acceleration of the wheel, $a_t=10.3\ m/s^2$

It is assumed to find the angular acceleration of the wheel. It is given by :

$a_t=\alpha \times r$

Where

$\alpha$ is the angular acceleration of the wheel

So,

$\alpha =\dfrac{a_t}{r}$

$\alpha =\dfrac{10.3\ m/s^2}{0.2\ m}$

$\alpha =51.5\ rad/s^2$

So, the angular acceleration of the wheel is $51.5\ rad/s^2$. Hence, this is the required solution.

Answer 2

Answer:

51.5 rad/s^2

Explanation:

radius, r = 0.2 m

tangential speed, v = 50 m/s

tangential acceleration, a = 10.3 m/s^2

time, t = 3.10 s

The relation between the radial acceeration and the tangential acceleration is given by

a = r x α

where, α is the radial acceleration and r be the radius of the circular path

10.3 = 0.2 x α

α = 51.5 rad/s^2

Thus, the radial acceleration is 51.5 rad/s^2.