Question

You are swinging a bucket in a circle at a velocity of 7.8 ft/s. The radius of the circle you are making is 1.25 ft. The acceleration is equal to one over the radius multiplied by the velocity squared.
a. What is the acceleration of the bucket?
b. What is the velocity if the acceleration is 25 ft/sec2?

Answer 1

Given that,

velocity, v = 7.8 ft/s

Radius, r = 1.25 ft

To find,

a. What is the acceleration of the bucket?

b. What is the velocity if the acceleration is 25 ft/sec²?

Solution,

(a) The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

Put v = 7.8 ft/s and r = 1.25 ft

$a=\dfrac{(7.8)^2}{1.25}\\\\=48.672\ ft/s^2$

(b) Put a = 25 ft/s² to find velocity.

$v=\sqrt{ar} \\\\v=\sqrt{25\times 1.25} \\\\v=5.59\ ft/s$

Hence, this is the required solution.