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mistermangos
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Hello everyone, I have been working on a problem and I am having a hard time figuring out if this is the right answer.
A solid sphere of mass 10 kg and radius 10 cm is rotating about its axis.
A. Find its moment of inertia
B. What is the kinetic energy of the sphere if it has an angular velocity of 10 revolutions per minute?
C. A tangential force of 100N is applied for 10 seconds. How much will the sphere slow down if it starts with the same angular velocity as part B?
Moment of inertia of solid sphere = I = [itex]\frac{2}{5}[/itex]*MR^2
Rotational Kinetic Energy = [itex]\frac{1}{2}[/itex]I[itex]\omega[/itex]^2
A. I = 2/5 (10)(.1)^2 = 2/50 = .04 kg [itex]\bullet[/itex] m^2
B. [itex]\frac{10}{60}[/itex] rev/s = [itex]\frac{1}{6}[/itex] rev/s = [itex]\frac{\pi}{3}[/itex]
Rot. Ke = 1/2 (.04)([itex]\pi[/itex]/3)^2 = .0219 J
C. A tangential force is applied, so the equivalent torque is [itex]\tau[/itex] = Fr = (100)(.1) = 10
[itex]\tau[/itex] = I[itex]\alpha[/itex], so [itex]\alpha[/itex] = 10/.04 = -250 rad/sec^2 (negative because it has a clockwise direction I guess, since its asking for the slowdown)
[itex]\omega[/itex](t) = w[itex]_{0}[/itex] + [itex]\alpha[/itex]t
[itex]\omega[/itex](10) = [itex]\frac{\pi}{3}[/itex] + (-250)(10)[itex]\omega[/itex](10) = -2498.95 rad/sec, or -397.72 rev/sec
seems a little fast for me... any help is appreciated
Homework Statement
A solid sphere of mass 10 kg and radius 10 cm is rotating about its axis.
A. Find its moment of inertia
B. What is the kinetic energy of the sphere if it has an angular velocity of 10 revolutions per minute?
C. A tangential force of 100N is applied for 10 seconds. How much will the sphere slow down if it starts with the same angular velocity as part B?
Homework Equations
Moment of inertia of solid sphere = I = [itex]\frac{2}{5}[/itex]*MR^2
Rotational Kinetic Energy = [itex]\frac{1}{2}[/itex]I[itex]\omega[/itex]^2
The Attempt at a Solution
A. I = 2/5 (10)(.1)^2 = 2/50 = .04 kg [itex]\bullet[/itex] m^2
B. [itex]\frac{10}{60}[/itex] rev/s = [itex]\frac{1}{6}[/itex] rev/s = [itex]\frac{\pi}{3}[/itex]
Rot. Ke = 1/2 (.04)([itex]\pi[/itex]/3)^2 = .0219 J
C. A tangential force is applied, so the equivalent torque is [itex]\tau[/itex] = Fr = (100)(.1) = 10
[itex]\tau[/itex] = I[itex]\alpha[/itex], so [itex]\alpha[/itex] = 10/.04 = -250 rad/sec^2 (negative because it has a clockwise direction I guess, since its asking for the slowdown)
[itex]\omega[/itex](t) = w[itex]_{0}[/itex] + [itex]\alpha[/itex]t
[itex]\omega[/itex](10) = [itex]\frac{\pi}{3}[/itex] + (-250)(10)[itex]\omega[/itex](10) = -2498.95 rad/sec, or -397.72 rev/sec
seems a little fast for me... any help is appreciated