Physics 131: Lecture 21. Today s Agenda


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1 Physics 131: Lecture 21 Today s Agenda Rotational dynamics Torque = I Angular Momentum Physics 201: Lecture 10, Pg 1
2 Newton s second law in rotation land Sum of the torques will equal the moment of inertia times the angular acceleration = An unbalanced net torque will result in an angular acceleration Physics 201: Lecture 10, Pg 2
3 Disk and box A box of mass 40 kg is tied to a light string wound around a wheel that has a mass of 30 kg and radius 0.5 m. Find the acceleration of the box. a =? Physics 201: Lecture 10, Pg 3
4 Clicker Question 1: A box of m B = 40 kg is tied to a light string wound around a wheel that has a m W = 30 kg and radius 0.5 m. Which is the correct equation of motion for the box? (a) T = m B g + m B a (b) T = m B g  m B a () (c) T = m B g (d) T = m W g  m W a (e) T=m B g  m W a a =? Physics 201: Lecture 10, Pg 4
5 Clicker Question 2: A box of m B = 40 kg is tied to a light string wound around a wheel that has a m W = 30 kg and radius 0.5 m. Which is the correct equation of motion for the wheel? (a) TR = m W R 2 (b) 2TR = (1/2)m W R 2 () (c) TR = (1/2)m W 4R 2 (d) TR = (3/4)m W R 2 (e) TR = (1/2)m W R 2 a =? Physics 201: Lecture 10, Pg 5
6 Clicker Question 3: A box of m B = 40 kg is tied to a light string wound around a wheel that has a m W = 30 kg and radius 0.5 m. How do the angular acceleration of the wheel and the linear acceleration a of the box compare? (a) (b) a = R a= 2R (c) a = 4R (d) a = R/2 a =? (e) a = R/4 Physics 201: Lecture 10, Pg 6
7 Constraints Due to Ropes and Pulleys A rope passes over a pulley and is connected to an object in linear motion. The rope does not slip as the pulley rotates. Tangential velocity and acceleration of the rim of the pulley must match the motion of the object: Physics 201: Lecture 10, Pg 7
8 Disk and box Now we remember a T = R So we can relate and a a =? a 1 2 m B g m m W B Physics 201: Lecture 10, Pg 8
9 Clicker Question 4: Block A slides upon a frictionless table, as shown. It is connected to a string, the other end of which is connected to a hanging block B. The string goes over a frictionless, massive pulley. As the two blocks move, the string does not slip on the pulley. At the moment shown in the figure, which of the following statements is true about the tension in the top part of the string, T 1, the tension in the lower part of the string, T 2, and the force of gravity on block B, mg? (a) T 1 = T 2 = mg (b) T 1 = T 2 < mg (c) T 1 < T 2 = mg (d) T 1 = T 2 > mg (e) T 1 < T 2 < mg Physics 201: Lecture 10, Pg 9
10 Clicker Question 4: Physics 201: Lecture 10, Pg 10
11 Course Evaluations I believe you ve been contacted by concerning online course evaluations These are very important in helping us improve the course experience Please ensure that students are checking their utoronto.ca address and not a departmental t or other address, and that t they check their spam folder, as s have sometimes been directed to those folders. Physics 201: Lecture 10, Pg 11
12 Rolling A wheel is spinning clockwise such that the speed of the outer rim is 2 m/s. What is the velocity of the top of the wheel relative to the ground? What is the velocity of the bottom of the wheel relative to the ground? 2 m/s y x 2 m/s Physics 201: Lecture 10, Pg 12
13 Rolling You now carry the spinning wheel to the right at 2 m/s. What is the velocity of the top of the wheel relative to the ground? A) 4 m/s B) 2 m/s C) 0 m/s D) +2m/s E) +4 m/s What is the velocity of the bottom of the wheel relative e to the ground? A) 4 m/s B) 2 m/s C) 0 m/s D) +2m/s E) +4 m/s 2 m/s 2 m/s V CM = 2m/s Physics 201: Lecture 10, Pg 13
14 Rolling without slipping The linear velocity is the same as the tangential velocity on the edge of the rolling object So we can use the viewer equation v CM = r Physics 201: Lecture 10, Pg 14
15 Rolling Kinetic Energy When a disk rolls we assume it does not slip and the outside of the ball moves with a velocity such that: v = r KE TOT = ½mv 2 + ½ 2 =v/r = 1/2 mr KE TOT MV I 2 2 KE TOT V MV R V MR MV MV KE MV Physics 201: Lecture 10, Pg 15
16 Conceptual Problem When a hollow sphere rolls without slipping with a velocity of v, what kinetic energy does it have? KE = TOT ½mv ½ =v/r = 2/3 mr 2 KE TOT KE TOT MV I V MV MR MV R 2 MV 3 2 KE 5 MV 6 2 Physics 201: Lecture 10, Pg 16
17 Clicker Question 5: A hollow cylinder of mass m rolls down an incline of height 2 m. We want to find the velocity of its center of mass when it reaches the bottom of the incline. Since it s rolling without slipping how would we represent it s total KE? (a) (3/6)mv 2 (b) (3/5)mv 2 (c) (7/6)mv 2 (d) mv 2 2 m (e) (3/4)mv 2 Physics 201: Lecture 10, Pg 17
18 Clicker Question 6: A hollow cylinder of mass m rolls down an incline of height 2 m. What velocity will its center of mass have at the bottom of the incline? (a) 9.0 m/s (b) 7.13 m/s () (c) m/s (d) 4.0 m/s (e) 56m/s m Physics 201: Lecture 10, Pg 18
19 Clicker Question 6: A hollow cylinder of mass m rolls down an incline of height 2 m. What velocity will its center of mass have at the bottom of the incline? 2m Physics 201: Lecture 10, Pg 19
20 Clicker Question 7: Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other. If both are placed at the top of the same ramp and released, which is moving faster at the bottom? (a) bigger one (b) smaller one (c) same Physics 201: Lecture 10, Pg 20
21 Clicker Question 8: Consider the following three objects, each of the same mass and radius: (1) a solid sphere (2) a solid disk (3) a hoop All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. In which order do the objects reach the bottom of the incline? (a) 1, 2, 3 hoop = mr 2 (b) 2, 3, 1 SS = 2/5 mr 2 (c) 3, 1, 2 (d) 3, 2, 1 SD = ½ mr 2 (e) All three reach the bottom at the same time. Physics 201: Lecture 10, Pg 21
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