Question

A tire placed on a balancing machine in a service station starts from rest and turns through 4.0 rev in 1.0 s before reaching its final angular speed. Assuming that the angular acceleration of the wheel is constant, calculate the wheel’s angular acceleration. Answer in units of rad/s 2 .

Answer 1

Answer:

a  = 50.26 rad/s^2

Explanation:

We know that:

θ = $\frac{1}{2}at^2$

where θ is the angle, a the angular aceleration and t the time.

First, we need to find how many rad are equivalent to 4 rev, as:

θ = 4 rev * 2π = 25.13 rad

Finally, replacing θ by 25.13 rad and t by 1 second, we get:

25.13 rad = $\frac{1}{2}a(1s)^2$

Solving for a:

a  = 50.26 rad/s^2

Answer 2

Answer:

50.27rad/s2

Explanation:

Using the formula,

= t + 1/2t2

Where = angular displacement of the rotating body = angle turned through by the body in rad, = initial angular velocity of rotating body in rad/s, =angular acceleration of the rotating body in rad/s2, t = time in s.

Since the body starts from rest, = 0, so the equation becomes,

= 1/2t2

Then,

= 2/t2,

Where t = 1s,

But,

for I complete revolution (rev) = 360 degrees = 2π,

Therefore, for 4 revs,

= 4 x 2π = 4 x 6.283185 = 25.1327rad

Substituting,

= 2 x 25.1327/1 x 1 = 50.27rad/s2