a. 170 rad/s^2 b. 1,63 rev how to solve this problems? N Some people are able to

Question

a. 170 rad/s^2

b. 1,63 rev

how to solve this problems?

N 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 18.65 rad/s in 1.10 s. a. Assuming the ball is subject to a constant angular acceleration, what is the magnitude of the constant acceleration? b. Through how many revolutions does the ball rotate during the 1.10 s?

Explanation / Answer

given that

initial angular velocity w1=0

finial angular velocity w2 =18.65

change time t=1.10 sec

(a)

this problem belong to circular motion of particle

therefore the angular acceleration = N {w2-w1/t}

the number of peoples N=10

the angular acceleration = 10{18.65-0}/1.10

=10*18.65/1.10

=169.5 rev/s^2

=170 rev/s^2

(b) ans

angular velocity w=4*pi*n/t

n:number of revolutions

n=wt/4*pi

=18.65*1.10/4*3.14

=20.515/12.36

=1.63 rev

therefore the angular acceleration =170 rev/s^2

the number of revolutions =1,63 rev

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