Mechanics Lecture 17, Slide 1 Physics 211 Lecture 17 Today’s Concepts: a) Torque Due to Gravity b) Static Equilibrium Next Monday 1:30-2:20pm, here: Hour Exam TWO Lect 08 to 13

First, a quick finish of ystrdy’s Rot Dynamics a2a2 T1T1 a1a1 T2T2  R M m1m1 m2m2 Mechanics Lecture 16, Slide 2 A massive pulley with weights attached like this is called an “Atwood’s machine”.

Atwood’s Machine a2a2 T1T1 a1a1 T2T2  R M m1m1 m2m2 Mechanics Lecture 16, Slide 3 Notice the similarity with this arrangement from Linear Motion.

Suppose a cylinder (radius R, mass M ) is used as a pulley. Two masses ( m 1  m 2 ) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. Compare the magnitudes of the acceleration of the two masses: A) a 1  a 2 B) a 1  a 2 C) a 1  a 2 ACT a2a2 T1T1 a1a1 T2T2  R M m1m1 m2m2 Mechanics Lecture 16, Slide 4 Demo: Atwood’s machine w/ massive pulley

Suppose a cylinder (radius R, mass M ) is used as a pulley. Two masses ( m 1  m 2 ) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. Compare the magnitudes of the acceleration of the two masses: A) a 1  a 2 B) a 1  a 2 C) a 1  a 2 ACT a2a2 T1T1 a1a1 T2T2  R M m1m1 m2m2 Mechanics Lecture 16, Slide 5 Demo: Atwood’s machine w/ massive pulley

ACT Suppose a cylinder (radius R, mass M ) is used as a pulley. Two masses ( m 1  m 2 ) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. How is the angular acceleration of the wheel related to the linear acceleration of the masses?: A)   Ra B)   a/R C)   R/a Mechanics Lecture 16, Slide 6 a2a2 T1T1 a1a1 T2T2  R M m1m1 m2m2

ACT Suppose a cylinder (radius R, mass M ) is used as a pulley. Two masses ( m 1  m 2 ) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. Compare the tension in the string on either side of the pulley: A) T 1  T 2 B) T 1  T 2 C) T 1  T 2 Mechanics Lecture 16, Slide 7 a2a2 T1T1 a1a1 T2T2  R M m1m1 m2m2

ACT Suppose a cylinder (radius R, mass M ) is used as a pulley. Two masses ( m 1  m 2 ) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley. Compare the tension in the string on either side of the pulley: When we have had a massless pulley, then we have been able to say T1 = T2 = T. But now our “pulley” is no longer massless. Mechanics Lecture 16, Slide 8 a2a2 T1T1 a1a1 T2T2  R M m1m1 m2m2

Atwood's Machine with Massive Pulley: A pair of masses are hung over a massive disk-shaped pulley as shown. Ø Find the acceleration of the blocks. For the hanging masses use F  ma  m 1 g  T 1  m 1 a  m 2 g  T 2  m 2 a For the pulley use   I  T 1 R  T 2 R (Since for a disk) Mechanics Lecture 16, Slide 9 y x m2gm2g m1gm1g a T1T1 a T2T2  R M m1m1 m2m2

We have three equations and three unknowns ( T 1, T 2, a ). Solve for a.  m 1 g  T 1   m 1 a (1)  m 2 g  T 2  m 2 a (2) T 1  T 2 (3) Mechanics Lecture 16, Slide 10 y x m2gm2g m1gm1g a T1T1 a T2T2  R M m1m1 m2m2 Atwood's Machine with Massive Pulley:

Mechanics Lecture 17, Slide 11 New Topic, Old Physics: As the name implies, “statics” is the study of systems that don’t move. Ladders, sign-posts, balanced beams, buildings, bridges... The key equations are familiar to us: If an object doesn’t move: The net force on the object is zero The net torque on the object is zero (for any axis) Statics or Rotational Equilibrium:

Mechanics Lecture 17, Slide 12 Statics: Example: What are all of the forces acting on a car parked on a hill? x y  N f mg

Mechanics Lecture 17, Slide 13 x y  N f mg Car on Hill: Use Newton’s 2nd Law: F NET  MA CM  0 Resolve this into x and y components: x  f  mg sin  0 f  mg sin  y  N  mg cos   0 N = mg cos Demo

Mechanics Lecture 17, Slide 14 Magnitude:  R CM Mg sin  Torque Due to Gravity Lever arm

Mechanics Lecture 17, Slide 15 Example: Now consider a plank of mass M suspended by two strings as shown. We want to find the tension in each string: L/2 L/4 M x cm T1T1 T2T2 Mg y x Demo: suspended static beam

Mechanics Lecture 17, Slide 16 Balance forces: T 1  T 2  Mg L/2 L/4 M x cm T1T1 T2T2 Mg y x

Mechanics Lecture 17, Slide 17 L/2 L/4 M x cm T1T1 T2T2 Mg y x Balance Torques Choose the rotation axis to be out of the page through the CM: The torque due to the string on the right about this axis is: The torque due to the string on the left about this axis is: Gravity acts at CM, so it exerts no torque about the CM  2  T 2 L/4  1  T 1 L/2

Mechanics Lecture 17, Slide 18 Finish the problem The sum of all torques must be 0 : We already found that T 1  T 2  Mg L/2 L/4 M x cm T1T1 T2T2 Mg y x

Mechanics Lecture 17, Slide 19 L/2 L/4 M x cm T1T1 T2T2 Mg y x What if you choose a different axis? Choose the rotation axis to be out of the page at the left end of the beam: The torque due to the string on the right about this axis is: The torque due to the string on the left is zero Torque due to gravity:

Mechanics Lecture 17, Slide 20 You end up with the same answer! The sum of all torques must be 0 : We already found that T 1  T 2  Mg Same as before L/2 L/4 M x cm T1T1 T2T2 Mg y x

Mechanics Lecture 17, Slide 21 Approach to Statics: Summary In general, we can use the two equations to solve any statics problem. When choosing axes about which to calculate torque, we can sometimes be clever and make the problem easier....

Mechanics Lecture 17, Slide 22 In Case 1 one end of a horizontal massless rod of length L is attached to a vertical wall by a hinge, and the other end holds a ball of mass M. In Case 2 the massless rod holds the same ball but is twice as long and makes an angle of 30 o with the wall as shown. In which case is the total torque about the hinge biggest? A) Case 1B) Case 2C) Both are the same gravity CheckPoint Case 1 Case 2 L 90 o M 2L2L 30 o M

Mechanics Lecture 17, Slide 23 C) Because sin(30) equals to 0.5. This makes up for the difference in the length of the rod. B) In case 2 the lever arm is larger than in case 1 so the torque would be larger. A) The force is perpendicular to the radius, meaning that cos(theta) is greatest, so torque is greater. CheckPoint Response In which case is the total torque about the hinge biggest? A) Case 1B) Case 2C) Both are the same Case 1 L 90 o M Case 2 2L2L 30 o M

Mechanics Lecture 17, Slide 24 Torque =  = R F sin  Torque 1 =  = (L) (Mg) sin 90 = MgL Torque2 =  = (2L) (Mg) sin 20 = MgL CheckPoint Response In which case is the total torque about the hinge biggest? A) Case 1B) Case 2C) Both are the same Case 1 L 90 o M Case 2 2L2L 30 o M

Your Comments Mechanics Lecture 17, Slide 25 Torque also depends upon F, the force. That “displacement” is also known as the “lever arm”. good ? ? ? No, no, no. Maybe you mean the torque caused by the forces in the hinge. In that case, tell me more. Doubling the angle does not double the sine of the angle; be careful with the details. ? ? ? Generally good answers; thanks.

Mechanics Lecture 17, Slide 26 An object is made by hanging a ball of mass M from one end of a plank having the same mass and length L. The object is then pivoted at a point a distance L/4 from the end of the plank supporting the ball, as shown below. Is the object balanced? A) A)Yes B) B) No, it will fall left C) C)No, it will fall right M M L L/4 gravity CheckPoint

Mechanics Lecture 17, Slide 27 An object is made by hanging a ball of mass M from one end of a plank having the same mass and length L. The object is then pivoted at a point a distance L/4 from the end of the plank supporting the ball, as shown below. Is the object balanced? A) A)Yes B) B) No, it will fall left C) C)No, it will fall right M M L L/4 gravity CheckPoint

Mechanics Lecture 17, Slide 28 How far to the right of the pivot is the center of mass of just the plank. A) L/4 B) L/2 C) 3L/4 M x ACT

Mechanics Lecture 17, Slide 29 x Is the object balanced? A) Yes B) No, it will fall left C) No, it will fall right M L/4 C) The center of mass is to the right of the L/4 dividing mark, so it will fall to the right. B) The mass on the left is greater than the mass on the right so the object will fall to the left. A) Mass M balanced at L/4 on left; Mass M balanced L/4 on right. So system balanced. M ACT

Mechanics Lecture 17, Slide 30 Hmmm,... Not entirely clear,... Is it any clearer now? good That’s brief. good But the masses are NOT equal on both sides. There is (3/4) M on the right and only (1/2) M on the left. How can it be balanced? But the mass is NOT equal on either side. I don’t think I understand the details. Where did these terms come from? Just tell me more. Any better now? “at equal distance”; THAT’S important Any better now? Summer school will do that; it’s like drinking from a fire hose! Do get some rest, tho’. Tell me more; I don’t know what this means

Mechanics Lecture 17, Slide 31 L/2 M In Case 1, one end of a horizontal plank of mass M and length L is attached to a wall by a hinge and the other end is held up by a wire attached to the wall. In Case 2 the plank is half the length but has the same mass as in Case 1, and the wire makes the same angle with the plank. In which case is the tension in the wire bigger? A) Case 1B) Case 2C) Both are the same CheckPoint gravity Case 1Case 2 L M

Mechanics Lecture 17, Slide 32 In which case is the tension in the wire bigger? A) Case 1 B) Case 2 C) Same L/2 M L M ACT

Mechanics Lecture 17, Slide 33 In which case is the tension in the wire bigger? A) Case 1 B) Case 2 C) Same L/2 M L M ACT

Your Comments Mechanics Lecture 17, Slide 34 What about the LOCATION of F? Ouch! I’m confused. No. ? ? ? Torque isn’t LOCATED at the edge. ? ? ? good no Hang in there; the week-end’s almost here good ? ? ?

Mechanics Lecture 17, Slide 35 Which equation correctly expresses the fact that the total torque about the hinge is zero? A) B) C) L Mg T  The L cancels out: L sin(  ) ACT

Mechanics Lecture 17, Slide 36 Use hinge as axis Mg T HxHx HyHy mg

Mechanics Lecture 17, Slide 37 mg Mg T HxHx HyHy y x

Mechanics Lecture 17, Slide 38 Solve for m m

Mechanics Lecture 17, Slide 39 Mg T HxHx HyHy mg Use hinge as axis Which of these things make T smaller?