A student sits on a rotating stool holding two 3.0 kg objects. When his arms are
ID: 1978000 • Letter: A
Question
A student sits on a rotating stool holding two 3.0 kg objects. When his arms are extended horizontally, the objects are .92 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus the stool is 3.0 kg.m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.30 m from the rotation axis.a) Find the new angular speed of the student.
b) Find the kinetic energy of the student before and after the objects are pulled in.
c) How do you account for the change in kinetic energy, if there is any?
Explanation / Answer
same question with different numbers: A student sits on a rotating stool holding two 2.0 kg objects.? When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation, and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg·m2 and is assumed to be constant. The student then pulls the objects horizontally to 0.20 m from the rotation axis. (a) Find the new angular speed of the student. Enter a number.1 rad/s (b) Find the kinetic energy of the student system before and after the objects are pulled in. before Enter a number.2 J after Enter a number.3 J answer: Use a conservation of momentum approach. Initially, the rotational inertia of the student + stool + weights system is: Inet1 = I + 2*m*d1^2 I is the rotational inertia of empty handed student + stool m is the mass of one of the barbels in his hand, assumed as point masses. The leading factor of 2 shows up because there is one in each hand. d1 is the initial positioning of each barbell This system is rotating at angular velocity omega1. Its angular momentum is thus: L = Inet1*omega1 When the student pulls in the barbells, the new rotational inertia is: Inet2 = I + 2*m*d2^2 Angular momentum after pulling in barbells: L = Inet2*omega2 Because no external torques act on the freely rotating system, ANGULAR MOMENTUM is conserved. Thus: Inet1*omega1 = Inet2*omega2 Solve for omega2: omega2 = omega1*Inet1/Inet2 And compute the rotational kinetic energies: KE1 = 1/2*Inet1*omega1^2 KE2 = 1/2*Inet2*omega2^2 Substitute omega2: KE2 = 1/2*Inet2*(omega1*Inet1/Inet2)^2 Simplify: KE2 = Inet1^2 * omega1^2/(2*Inet2) Substitute net rotational inertia expressions: omega2 = omega1*( I + 2*m*d1^2)/(I + 2*m*d2^2) KE1 = 1/2*(I + 2*m*d1^2)*omega1^2 KE2 = (I + 2*m*d1^2)^2 * omega1^2/(2*(I + 2*m*d2^2)) Data: omega1:=0.75 rad/second; I:=3.0 kg-m^2; m:=2 kg; d1:=1 m; d2:=0.2 m; Results: A) omega2 = 1.661 radians/second B) KE1 = 1.969 Joules KE2 = 4.361 Joules Checkpoint answers are incorrect. So, we used conservation of angular momentum to solve this problem... But why is kinetic energy not conserved? Where did this KE come from? Well, the student needed to do 2.392 Joules of work with his muscles to bring the barbels inward toward the axis of revolution.
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