Amerry-go-round in the park can be considered a circular disk ofmass 100 kg and

Question

Amerry-go-round in the park can be considered a circular disk ofmass 100 kg and a radius of 2.0 m. A student whose mass is 60 kgruns toward the merry-go-round in a direction that is tangent toits edge at a speed of 7.0 m/s. The merry-go-round is initially atrest. The student can be considered a point mass. I(disk) =½ MR2.

a.       A.The student jumps on theedge of the merry-go-round. What is the resulting angularvelocity of the merry-go-round?

b.     B. The student then slowly walks toward the center of themerry-go-round, and stops 0.5 m from the center. What is theresulting angular velocity?

c.       C. A second studentthen steps up and stops the merry-go-round by applying a steadyforce tangentially to the rim for 20 s. Calculate theresulting angular acceleration.

d.     D.How much force is required by the second student to stop themerry-go-round in this manner?

Explanation / Answer

Data v = 7m/sec                      speedof the student r = 2m                             radiusof the merry go round m = 60kg                         massof student M = 100kg                      massof merry go round A) The angular moment is conserved before and after the studentjumps. Before the student jumps, the merry-go-round is at rest, andafter the jump, both are moving with the same angular speed1 L0 = Lf I00 =I01+Im1 where I0 and 0 are the momentof inertia of the student, and shi angular speed, withrespect to the center of the merry-go-round (the rotationaxis). Im is the moment of inertia of the mgr. The aboveequation can be rearranged as 1 = (I0/ ( I0 +Im))0                (1) We also have the following three equations for the angularspeed and the moments of inertia 0  = v/r I0 = mr2 Im = (1/2)Mr2 Substituing in equation (1) yields 1 = 2mv/( ( 2m + M) r ) 1 = 1.9090 rad/sec B) The angular moment when the student is at distance r from thecenter, is the same as when he is at r2 = 0.5 m, becausehe walks slowly. Therefore (I0 + Im) 1 =(I1 +Im)2              (2) where I0 = mr2 Im = (1/2)Mr2 I0 = mr2 Im = (1/2)Mr2 I1 = mr22 We want to find the value of 2. From equation(2), and using the above three equation yields [ (I0 + Im) /(I1 +Im) ]1=2    [ (I0 +Im) /(mr22 +Im) ]1= 2 2 = 3.906976rad/sec 2 = 3.906976rad/sec C) Since the force applied is steady, the angular accelerationwill be constant. Therefore, its equation is = (final - initial)/t = ( 0 - 2)/t = - 2/t where t = 20 sec. Then = 3.9069/20 = -0.19534rad/sec2
D) The force made by thestudent generates a torque. We assume that the first student isstill at r2 = 0.5 m from the center of the mgr.Then T =Fr By the second law ofnewton Fr = (Im+I1) F = (Im+I1) / r F = -21N

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