(a) The student is given an initial angular speed while holding two weights out.

Question

(a) The student is given an initial angular speed while holding two weights out. (b) The angular speed increases as the student draws the weights inwards.

Goal Apply conservation of angular momentum to a simple system.

Problem A student sits on a pivoted stool while holding a pair of weights. (See Figure) The stool is free to rotate about a vertical axis with negligible friction. The moment of inertia of student, weights, and stool is 2.25 kg · m2. The student is set in rotation with arms outstretched, making one complete turn every 1.26 s, arms outstretched. (a) What is the initial angular speed of the system? (b) As he rotates, he pulls the weights inward so that the new moment of inertia of the system (student, objects, and stool) becomes 1.80 kg · m2. What is the new angular speed of the system? (c) Find the work done by the student on the system while pulling in the weights. (Ignore energy lost through dissipation in his muscles.)

Strategy (a) The angular speed can be obtained from the frequency, which is the inverse of the period. (b) There are no external torques acting on the system, so the new angular speed can be found with the principle of conservation of angular momentum. (c) The work done on the system during this process is the same as the system's change in rotational kinetic energy.

SOLUTION

(a) Find the initial angular speed of the system.

Invert the period to get the frequency, and multiply by 2.

i = 2f = 2/T = 4.99 rad/s

(b) After he pulls the weights in, what's the system's new angular speed?

Equate the initial and final angular momenta of the system.

(1)

Li = Lf Iii = Iff

Substitute and solve for the final angular speed f.

(2)

(2.25 kg · m2)(4.99 rad/s) = (1.80 kg · m2)f

f = 6.24 rad/s

(c) Find the work the student does on the system.

Apply the work—energy theorem.

Wstudent = Kr = 1/2Iff2 1/2Iii2

= 1/2(1.80 kg · m2)(6.24 rad/s)2 1/2(2.25 kg · m2)(4.99 rad/s)2

Wstudent = 7.03 J

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Remarks Although the angular momentum of the system is conserved, mechanical energy is not conserved because the student does work on the system.

Question If the student suddenly releases the weights, how does his own angular momentum, rotational kinetic energy, and angular speed change? (Select all that apply.)

His rotational angular speed remains the same. His angular momentum increases. His rotational kinetic energy remains the same. His angular momentum remains the same. His rotational angular speed decreases. His angular momentum decreases. His rotational angular speed increases. His rotational kinetic energy increases. His rotational kinetic energy decreases.

EXERCISE HINTS:  GETTING STARTED | I'M STUCK!

Tafter = s

Wcollapse = J

(c) What is the speed of an indestructible person standing on the equator of the collapsed star? (Neglect any relativistic or thermal effects, and assume the star is spherical before and after it collapses). vequator = m/s

Explanation / Answer

Net torque is zero. so Angular momentum is constant

From conservation of angualr momentum

L =Iw =constnat

I =mr^2, Moment of inertia decresed, so angular speed is incresed.

K=L^2/2I

I =mr^2, Moment of inertia decresed, so rotationla kinetic energy is incresed.

R1 =1x10^8 m , T1 =30 days , R2 =1x10^4 m

(a) From Keplers third law

T2 propotional to R3

(T2/T1)2 =(R2/R1)3

T2 = T1 (R2/R1)3/2= (30days)(104/108)3/2

T2 = 30x10-6 days =30x24x60x60x10-6 s

time period after collapse T2 =2.592s

(b) Work done by gravity is change in potential energy. Because gravitaiotnal force is conservative force

m = 2x10^30 kg, Mass of earth M =5.98x10^24 kg

G =6.67x10^-11 N.m^2/kg^2

W =Ui -Uf =- (GMm/R1) + (GMmR2) = GMm[(1/R2) -(1/R1)]

= (6.67x10-11x2x1030x5.98x1024] [(1/104) -(1/108)]

W= 7.98x1040 J

(C) v =2p*R2/T2 = (2*3.14*1x104)/(2.592)

v =24 km/s

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