A clothes dryer undergoes an centripetal acceleration of 28m/s^2 if the dryer has a radius of 27cm what is the trangential speed of a sock struck to the outside of the dryers rotating barrel?
To find the tangential speed of the sock stuck to the outside of the dryer's rotating barrel, we need to use the formula:
tangential speed (v) = radius (r) * angular velocity (ω)
First, we need to find the angular velocity (ω). We are given the centripetal acceleration (a_c) of the dryer, which is the same as the acceleration due to the rotation. But we know that the centripetal acceleration is related to the angular velocity by the formula:
a_c = r * ω^2
where r is the radius and ω is the angular velocity.
We can rearrange this equation to solve for ω:
ω = √(a_c / r)
Given:
a_c = 28 m/s^2
r = 27 cm = 0.27 m
Plugging in the values, we get:
ω = √(28 / 0.27) = √(103.7) ≈ 10.18 rad/s
Now we can find the tangential speed (v) using the formula:
v = r * ω
Plugging in the values, we get:
v = 0.27 * 10.18 ≈ 2.75 m/s
Therefore, the tangential speed of the sock struck to the outside of the dryer's rotating barrel is approximately 2.75 m/s.