University Physics Quiz 7 Solution Fri Oct 25, 2012 Your name: A third problem is on the back of this sheet!!!! Please show your work, write neatly, and put a box or circle around each of your answers. This week the quiz is also being given in MasteringPhysics with the same problems. 1. (4 points total) Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. (a) What is the angular acceleration of the salad spinner as it slows down, in degrees/s2 ? (b) How long does it take for the salad spinner to come to rest, in seconds? Solution: The problem description gives a few of our angular variables. Its initial angular velocity is ω0 = (20 revs ∗ (360◦ /rev))/(5.0 s) = 1440◦ /s. Its final angular velocity is ω = 0◦ /s. It moves through a total angle of (θ − θ0 ) = (6.00 rev) ∗ (360◦ /rev) = 2160◦ as it’s slowing down. We then can use our “standard” angular kinematics equations to find α: ω 2 = ω02 + 2α(θ − θ0 ) ⇒ α= and ω = ω0 + αt ⇒ t= ω 2 − ω02 0 − (1440◦ /s)2 = = −480◦ /s2 = α 2(θ − θ0 ) 2(2160◦ ) ω − ω0 0 − (1440◦ /s) = = 3s=t α −480◦ /s2 2. (3 points total) A motor is connected to a solid cylindrical drum with a diameter of 1.2 m and a mass of 51 kg. A massless rope is attached to the drum and tied at the other end to a 38 kg weight, so the rope will wind onto the drum as it turns. (a) What is the rotational inertia of the drum? i. I=18.36 kg-m2 . ii. I=15.30 kg-m2 . iii. I=6.12 kg-m2 . iv. I=9.18 kg-m2 . correct choice v. I=1.53 kg-m2 . (b) What torque must the motor apply if the weight is to be lifted with an acceleration of 1.1 m/s2 ? Express your answer using two significant figures. Solution: The rotational inertia of a solid cylinder is I = (1/2)M R2 where M = 51 kg is the cylinder mass, and R = 1.2 m/2 = 0.6 m is the cylinder radius. Then we have 1 I = M R2 = 9.18 kg m2 = I 2 This is item (iv) in the multiple choice list above. The rest of the problem is similar to the “bucket in a well” problem that was done in class. Taking up as the positive direction, we can write Newton’s second law for the hanging weight of mass m = 38 kg that is accelerating at a = 1.1 m/s2 , with T as the tension in the rope: X F = T − mg = ma ⇒ T = m(g + a) = (38 kg)(10.9 m/s2 ) = 414 N 1 We can also write the angular form of Newton’s second law for the cylindrical drum, where α = a/r = (1.1 m/s2 /(0.6 m) = 1.83 rad/s2 : X τ = τapplied − T r = Iα rad 2 τapplied = Iα + T r = (9.18 kg m ) 1.83 2 + (414 N)(0.6 m) = 267 N m ≈ 270 N − m = τapplied s 3. (3 points total) A 57 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 7.0 g , moves along the x axis at 36 m/s in the positive direction. The second, with mass 19 g, moves along the y axis at 27 m/s in the positive direction. (a) This is a “boom” problem – the firecracker explodes! Can you use conservation of momentum, conservation of energy, both, or neither to solve this problem? i. ii. iii. iv. Only conservation of momentum. correct choice Neither conservation of energy nor conservation of momentum. Both conservation of energy and conservation of momentum. Only conservation of energy. (b) Find the x-component of the velocity of the third piece. (c) Find the y-component of the velocity of the third piece. Solution: Technically both momentum and energy are always conserved. But here you cannot use the conservation of energy to solve the problem because energy is added to the system by the explosion of the firecracker, and you have no straightforward way of calculating how big that energy is to make both sides of the energy conservation equation balance. So you can really only use the conservation of momentum, choice (i). We must then use the conservation of momentum to solve for each velocity. The leftover mass of the firecracker is M = 57 g − 7.0 g − 19 g = 31 g. In the x direction: X X px,init = px,final 0 = (7 g)(36 m/s) + (31 g)vx vx = −(7 g)(36 m/s)/((31 g) = −8.13 m/s = vx Then in the y direction: X X py,init = py,final 0 = (19 g)(27 m/s) + (31 g)vy vy = −(19 g)(27 m/s)/((31 g) = −16.5 m/s = vy 2