Hwk #9

Part A: You would first pick a pivot at the edge of the table since it would be the center when the remote does tip over. You would then add up the torques around the point. However, for mg radius it would be (Ltot/2-L) because of the fact that that is the radius. Solve for L from there. Part B: Use the Fnety = 0 and solve for the normal force.

A 0.100 kg remote control 16.7 cm long rests on a table, as shown in the figure below, with a length L overhanging its edge. To operate the power button on this remote requires a force of 0.395 N. Assume the mass of the remote is distributed uniformly, and that the power button is on the end of the remote overhanging the table. a.) How far can the remote control extend beyond the edge of the table and still not tip over when you press the power button? (Answer cm) b.) How large is the normal force when the button is pressed? (Answer N) (How would this answer change if the remote were not at maximum extension?)

Part A: You would use the center of mass formula in the x direction: M1*X1 + M2*X2 + etc./ M1 + M2 + etc. Part B-C: You would pick a pivot point and use the inertia formula: I = mr^2 + mr^2. Note: for one of the rs in the edges, it would be zero because that would be your pivot point. Part D: same thing but the radius would be r/2 for both since we are going to the center of the rod. Part E: Same thing but the pivot point is the COM Part F: You would need to use the Tnet equation: Tnet = Ia (I = Inertia and a = angular acceleration). From there you would find the Tnet by using the torque equation: Frsin(theta). Then you would find inertia and plug it in to find angular acceleration.

A 3 kg mass and a 6 kg mass are attached to either end of a 3 m long massless rod. a.) Find the center of mass of the system.(Answer m) , from the 3 kg mass. Find the rotational inertia (I) of the system when rotated about: b.) the end with the 3 kg mass.(Answer kg m^2) c.) the end with the 6 kg mass. (Answer kg m^2) d.) the center of the rod. (Answer kg m^2) e.) the center of mass of the system. (Answer kg m^2) (Compare this to parts b-d. Is this what you expect?) f.) Suppose a frictionless pivot is attached at the center of mass, such that the system is free to rotate about this point in the horizontal plane. A force of 10 N is exerted perpendicular to the rod, pushing directly on the 6 kg mass. What is the size of the system's angular acceleration that would result?. (Answer rad/s^2)

You would have to use the equation of Tnet = Ia (I = Inertia and a = angular acceleration) in order to find the angular acceleration. You would then need to find the inertia by using the inertia equation: I = mr^2.

A couterclockwise torque of +0.89 Nm is applied to a bicycle wheel of radius 26 cm and mass 0.63 kg. Treating the wheel as a thin ring, find its angular acceleration. (Answer rad/s^2)

Part A: Use the panel inertia equations: 1/12*M*l^2 for the full length panel (2) or 1/3*M*l^2 for the short panels (4) . Then you would add up Part B: Use the Tnet equation: Tnet = Ia (I = Inertia and a = angular acceleration). To find the Tnet, you would use torque equation: T= r*f*sin(theta). Use the inertia from part A and find angular acceleration. Part C: Use the constant acceleration in order to find time.

A rotating door is made from four rectangular glass panes, as shown in the drawing. The mass of each pane is 90 kg. A person pushes on the outer edge of one pane with a force of size 64 N and directed perpendicular to the pane. a.) Find the rotational inertia of the door. (Answer kg m^2) b.) Find the magnitude of the door's angular acceleration..(Answer rad/s^2) c.) Find the time it would take to rotate the door a half-circle, if it started from rest. (Answer s)

Part A: you would need the Tnet equation in order to find angular acceleration: Tnet = Ia (I = Inertia and a = angular acceleration). Then you would need to find Inertia by using the solid hole circle inertia formula: .5*m*r^2. You could also derive it by using the net inertia and adding up the two separate inertias. Then, you would need to find Tnet = Frsin(theta). Part B: Use the constant angular acceleration equations to find for delta theta. With the delta theta, you would need to put it displacement equation: s= r * delta(theta). Part C: two options, do everything again with the radius halved or use proportionalities.

Approximate a fishing reel to be two solid disks, each of radius 5.6 cm and combined mass 0.87 kg which can rotate without friction. String (of negligible mass) is wrapped around the reel (between the disks) all the way to its edge (see Fig 1a). The fishing line is strung with bait and left at rest in the water. A fish then suddenly takes the bait and applies a constant force of 1.9 N on the fishing line, pulling the reel counterclockwise. (a) What is the angular acceleration of the fishing reel?(Answer rad/s^2) (b) Immediately after taking the bait, how much fishing line does the fish pull from the reel in 0.19 s? (Answer s) (Assume this time interval is short enough that radius of the fishing line wrapped around the reel does not appreciably change during it.) (c) Suppose instead this fisherman had cast his bait much further out, so that the fishing line left on the reel was only halfway out to the edge (see Fig 1b). If all other givens are the same, re-answer parts a & b. (Answer rad/s^2) (Answer s)

Part A: It has to be larger than the combined weight because of the fact that in order to throw the ball, it would need to lift the arm and the ball in the first place. Because of this fact, it would have to be greater. Part B: First you need to find the Tnet and Fnet of each force and replace each torque force in the Tnet with the force * radius from the chosen pivot point. From there, just find the force of the bicep. Part C:From the Fnet, add up the forces that are set equal to 0 and find the upper arm bone force. It would also be downwards because of the fact that it needs to cancel the force of the bicep.

Below is a diagram of a baseball/forearm system at rest. It is acted on by four different forces: the weight of the forearm, the weight of the baseball, the bicep force, and a force from the upper arm bone (attached at the elbow). Use the following values: length of forearm = L = 13 cm elbow to bicep distance = d = 2 cm mass of forearm = M = 3 kg mass of baseball = m = 1 kg a.) Based on torque considerations, the force of the bicep must be (Answer direction) the combined weight of the baseball/forearm.(Make sure you understand the reasoning.) b.) How large is the force exerted by the bicep?(Answer N) c.) The upper arm bone exerts a force (not shown above) on the forearm. This force must be: size: (Answer N), dir: (Answer direction)

Part A:Use the constant acceleration equations for angular acceleration in order to find the angular acceleration. In order to get delta theta, you would need to see the relationship of 1 revolution = 2*pi. From there, you would plug it in the Tnet equation: Tnet = Ia (I = Inertia and a = angular acceleration). Part B: Depends on the what is doubled and what is halved. From there, you would use proportionalities to find whether or not it increase or decreases.

On the Wheel of Fortune game show, a solid wheel (of radius 7.5 m and mass 10 kg) is given an initial counterclockwise angular velocity of +1.14 rad/s. It then smoothly slows down and stops after rotating through 3/4 of a turn. (a) Find the frictional torque that acts to stop the wheel. (Answer N*m) (b.)Assuming that the torque is unchanged, but the mass of the wheel is halved and its radius is doubled. How will this affect the angle through which it rotates before coming to rest? (Make sure you can reason this with proportionalities.) increase decrease stay the same

You first need to find the Fnet of y and the Torque net (all of the forces added up) from your chosen pivot point. Fnety = 0 and Tnet = 0. They both go to zero because the object is not in motion, but rather is stationary with all these forces. From there, you would use the Tnet equation to find a force by replacing the each torque force with Force * radius from pivot point. Then find the other force from the Fnet.

The object shown below (length = 10 m) remains completely stationary, despite the fact that three forces act on it. Forces 1 and 3 act at either end, while force 2 acts 1/4 of the way from the left end. If the size of force 3 is 4 N, then how large must the other two forces be? F1 = (Answer N) F2 = (Answer N) (For excellent practice, see if you can also get these answers using different pivot

Part A: you need to find the Fnet in order to find the students mass. Part B: You would need to first pick a pivot point in order to find the Tnet. Then you would need to find all the torque forces by using the torque formula: r*F*sin(theta) for each force. Note: for your pivot point, the torque force there would be zero because it is not turning.

To determine the location of his center of mass, a physics student lies on a lightweight plank supported by two scales L = 2.90 m apart. (a) If the left scale reads 349 N, and the right scale reads 143 N, find the student's mass.(Answer kg) (b) Find the distance from the student's head to his center of mass. (Answer m)

Part A: Use the center of mass formula for the x direction: M1*X1 + M2*X2 + etc./ M1 + M2 + etc. Part B: You would attach the string to the center of mass because then it would hold the object in equal in order for it to be completely horizontal. Part C: Pick the center of mass and use the torque formula: T= r*f*sin(theta). It would also be downwards since there is more force on the left side without the weight. Part D: You would use the Tnet equation: Tnet = Ia (I = Inertia and a = angular acceleration).

Two masses (mA= 1 kg, mB= 5 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively. a.) Where is the center of mass of this system? (Answer cm), from mass A. b.) The system is then hung from a string, so that it stays horizontal. Where should the string be placed? (Answer cm), from mass A. c.) Now, if mass B was removed, how much force would need to be exerted at the 100 cm mark in order to keep the meter stick level*?size: (Answer N), dir: (Answer direction) d.) Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant*. (Answer rads/s^2) *For parts c-d, assume the string remains attached at the same location you found in part b.