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midterm 04 CC10-solutions

# midterm 04 CC10-solutions - Version 047/AACDD midterm 04...

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Version 047/AACDD – midterm 04 CC10 – turner – (57340) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uniform solid disk with radius R and mass M is mounted on a fixed axle passing through its center (which is not shown). A constant force F is applied as shown. At the instant shown, the angular velocity of the disk ω is in the - ˆ z direction (where +ˆ x is to the right, +ˆ y is up, and +ˆ z is out of the page, toward you). The length of the string is d. The angular momentum, torque and kinetic energy of the disk are respectively: 1. - 1 2 MR 2 ω ˆ z , RF ˆ z , - 1 4 MR 2 ω 2 2. 1 2 MR 2 ω ˆ z , RF ˆ z , - 1 4 MR 2 ω 2 3. 1 2 MR 2 ω ˆ z , - RF ˆ z , - 1 4 MR 2 ω 2 4. 1 2 MR 2 ω ˆ z , - RF ˆ z , 1 4 MR 2 ω 2 5. - 1 2 MR 2 ω ˆ z , - RF ˆ z , 1 4 MR 2 ω 2 correct 6. - 1 2 MR 2 ω ˆ z , RF ˆ z , 1 4 MR 2 ω 2 7. 1 2 MR 2 ω ˆ z , RF ˆ z , 1 4 MR 2 ω 2 8. - 1 2 MR 2 ω ˆ z , - RF ˆ z , - 1 4 MR 2 ω 2 Explanation: The direction of both L and thtat of torque are along the - ˆ z direction. The moment of inertia of the disk is given by I = 1 2 M R 2 The magnitude of angular momentum is L = I ω = 1 2 M R 2 ω The magnitude of the torque is given by τ = R 2 + d 2 F sin θ and the value of sin θ is given by sin θ = R R 2 + d 2 Thus, the value of torque is τ = R F The kinetic energy is given by K = 1 2 I ω 2 = 1 4 M R 2 ω 2 002 10.0 points Consider a ball with its mass 0.1 kg is head- on colliding with a lead-sphere. Initially the lead-sphere is at rest and the ball is moving along the positive x direction with a speed 6 m/s and the final velocity is along the negative x-direction with a speed 4 m/s. Assume the lead-sphere is so massive that its final kinetic energy is negligible compared to the kinetic energies of ball, both before and after the collision. Consider following choices for the change in the momentum vector of the lead- sphere Ia. Δ p lead - sphere = Δ p ball Ib. Δ p lead - sphere = 0 Ic. Δ p lead - sphere = - Δ p ball and the following choices for the change of the internal energy IIa. Δ E int = 1 J IIb. Δ E int = - 1 J IIc. Δ E int = 0 1. Ia, IIa 2. Ia, IIb 3. Ib, IIc 4. Ic, IIc

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Version 047/AACDD – midterm 04 CC10 – turner – (57340) 2 5. Ic, IIb 6. Ib, IIb 7. Ib, IIa 8. Ia, IIc 9. Ic, IIa correct Explanation: Let : m = 0 . 1 kg , v i = 6 m / s , and v f = 4 m / s We take the ball plus the lead-sphere as a system. Neglecting any external impulse, the momentum principle implies that Δ p ball + Δ p lead - sphere = 0 . So Ic is the correct choice. For the ball + lead-sphere system, the en- ergy principle states that Δ E tot = Δ K + Δ E int = W + Q . Since collision process occurs quickly, W and Q are assumed to be negligible. Thus Δ K + Δ E int = 0 . This leads to Δ E int = - Δ K ball - Δ K lead - sphere = - Δ K ball = K i - K f = m 2 v 2 i - m 2 v 2 f = (0 . 1 kg)((6 m / s) 2 - (4 m / s) 2 ) 2 = 1 J .
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midterm 04 CC10-solutions - Version 047/AACDD midterm 04...

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