Some people are able to spin a basketball on the tip of a finger. It is...

Some people are able to spin a basketball on the tip of a finger. It is often done by balancing the ball on the tip of the finger and brushing the other hand along the side of the ball to cause it to rotate. Suppose the ball begins from rest and reaches a final angular speed of 20.15 rad/s in 1.55 s.

(a) Assuming the ball is subject to a constant angular acceleration, what is the magnitude of the constant angular acceleration? rad/s2

(b) Through how many revolutions does the ball rotate during the 1.55 s?

Solutions

Expert Solution

Time period = T = 1.55 sec

Initial angular speed of the ball = ₁ = 0 rad/s

Angular speed of the ball after 1.55 sec = ₂ = 20.15 rad/s

Angular acceleration of the ball =

₂ = ₁ + T

20.15 = 0 + (1.55)

= 13 rad/s²

Angle through which the ball rotates in 1.55 sec =

= ₁T + T²/2

= (0)(1.55) + (13)(1.55)²/2

= 15.616 rad

Number of revolutions made by the ball in 1.55 sec = n

n = 2.485 rev

a) Magnitude of the constant angular acceleration = 13 rad/s²

b) Number of revolutions made by the ball in 1.55 sec = 2.485 rev

genius_generous answered 8 months ago

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