A ball with radius .15 m is rolling with an angular velocity of 6.5 rad/s. (a) What linear distance will the ball roll in 5.0 seconds? The ball then slows down with a linear acceleration of -1.2 rad/s2. (B) How fast will it be rolling rad/s after .65 second? (C) If a piece of gum is stuck on the outer part of the ball, what is its linear speed? (4.9, 5.72, .858)

(a) 4.875 m

(b) 5.72 rad/s

(c) 0.858 m/s

Explanation:

(a) Assuming constant angular speed, the angular distance the ball would have traveled after 5s at the rate of 6.5 rad/s is

6.5 * 5 = 32.5 rad

With radius of 0.15m, the linear distance it would have traveled is

32.5 * 0.15 = 4.875 m

(b)The angular velocity of the ball after 0.65s when subjected to an angular acceleration of -1.2 rad/s is

6.5 - 1.2*0.65 = 5.72 rad/s

(c)The linear speed of the ball is the product of the angular speed and radius 0.15 m

5.72 * 0.15 = 0.858 m/s

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .