Physics 2 Chapter 10 problems
10.6 A machinist is using a wrench to loosen a nut. The wrench is 25cm long, and he exerts a 17-N force at the end of the handle. a) What torque does the machinist exert about the center of the nut? b) What is the maximum torque he could exert with this force?
10.6 A machinist is using a wrench to loosen a nut. The wrench is 25cm long, and he exerts a 17-N force at the end of the handle. a) What torque does the machinist exert about the center of the nut? b) What is the maximum torque he could exert with this force? Use the definition of torque. The direction of the torque will be out of the page. r F r F sin( ) (0.25m) (17N) sin( 37 ) 2.56N m Maximum torque will always occur when the radius and force are perpendicular to each other max (0.25m) (17N) sin(90 ) 4.25N m
10.12 A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley. The pulley is a uniform disk with mass 10kg and radius 50cm and turns on frictionless bearings. You measure that the stone travels 12.6m in the first 3 seconds, starting from rest. Find a) the mass of the stone and b) the tension in the wire.
10.12 A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley. The pulley is a uniform disk with mass 10kg and radius 50cm and turns on frictionless bearings. You measure that the stone travels 12.6m in the first 3 seconds, starting from rest. Find a) the mass of the stone and b) the tension in the wire. Find acceleration from kinematics: Net Torque on pulley: Net Force on stone: T T Mg
10.20 A string is wrapped several times around the rim of a small hoop with radius 8cm and mass 0.18kg. The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 75cm, calculate a) the angular speed of the rotating hoop and b) the speed of its center.
10.20 A string is wrapped several times around the rim of a small hoop with radius 8cm and mass 0.18kg. The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 75cm, calculate a) the angular speed of the rotating hoop and b) the speed of its center. Use conservation of energy for this one:
10.26 A bowling ball rolls without slipping up a ramp that slopes upward at an angle to the horizontal. Treat the ball as a uniform, solid sphere, ignoring the finger holes. a) Draw the free-body diagram for the ball. Explain why friction is uphill. b) What is the acceleration of the center of mass of the ball? c) What minimum coefficient of static friction is needed to prevent slipping?
10.44 A solid wood door 1m wide and 2m high is hinged along one side and has a total mass of 40kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 5kg, traveling perpendicular to the door at 12m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?
10.62 A large 16kg roll of paper with radius R=18cm rests against the wall and is held in place by a bracket attached to the rod through the center of the roll. The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.26kg-m 2. The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of 30 with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is μ k =0.25. A constant vertical force F=40N is applied to the paper, and the paper unrolls. a) What is the magnitude of the force the rod exerts on the paper as it unrolls? b) What is the angular acceleration of the roll?
10.62 A large 16kg roll of paper with radius R=18cm rests against the wall and is held in place by a bracket attached to the rod through the center of the roll. The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.26kg-m 2. The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of 30 with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is μ k =0.25. A constant vertical force F=40N is applied to the paper, and the paper unrolls. N F rod a) What is the magnitude of the force the rod exerts on the paper as it unrolls? b) What is the angular acceleration of the roll? f k When the paper unrolls, the forces will all balance out, but the rotational motion means there will be a net torque on the roll. This gives us 2 force equations, and a torque equation to work with. mg Pivot point at rod, positive=clockwise rotation
10.75 A solid, uniform ball roll without slipping up a hill, as shown. At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff. a) How far from the foot of the cliff does the ball land, and how fast is it moving just before it lands? b) Notice that when the ball lands, it has a larger translational speed than when it was at the bottom of the hill. Does this mean that the ball somehow gained energy? Explain!
10.75 A solid, uniform ball roll without slipping up a hill, as shown. At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff. a) How far from the foot of the cliff does the ball land, and how fast is it moving just before it lands? b) Notice that when the ball lands, it has a larger translational speed than when it was at the bottom of the hill. Does this mean that the ball somehow gained energy? Explain! The ball will slow down as it rolls up the hill, so we need to find the speed at the top. Then this will turn into a projectile problem. We will use conservation of energy: Combine terms and cancel the m s This becomes the initial speed for the horizontal projectile. While the ball is free-falling back to the ground, its vertical component of velocity increases: The ball s speed decreases as it rolls up the hill, and when it free-falls back down, the translational kinetic energy increases, but Pythagorean theorem gets us the final speed: the ball doesn t spin faster (no torque). So total energy is the same, Prepared just by more Vince trans., Zaccone less rotational.