A ball rolling down a hill for 9 seconds accelerates from 3 m/s to 34.5 m/s. What is the ball's rate of acceleration?
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Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
<u>b) Determine the force exerted on B2 at P</u>
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
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