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A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle as in the figure below. A light cord wrapped around the wheel supports an object of mass m. When the wheel is released, the object accelerates downward, the cord unwraps off the wheel, and the wheel rotates with an angular acceleration. We apply two analysis models here. The object is modeled as a particle under a net force. The wheel is modeled as a rigid object under a net torque. The magnitude of the torque acting on the wheel about its axis of rotation is = TR, where T is the force exerted by the cord on the rim of the wheel. (The gravitational force exerted by the Earth on the wheel and the normal force exerted by the axle on the wheel both pass through the axis of rotation and therefore produce no torque.) Solve for a and substitute the net torque: ΣText = = Ια (1) α = Στοιχε ext I TR I Apply Newton's second law to the motion of the object, Σ Fy = mg - T = ma taking the downward direction to be positive: Solve for the acceleration a: (2) a mg - T = m Equations (1) and (2) have three unknowns: a, a, and T. Because the object and wheel are connected by a cord that does not slip, the translational acceleration of the suspended object is equal to the tangential acceleration of a point on the wheel's rim. Therefore, the angular acceleration a of the wheel and the translational acceleration of the object are related by a = Rα. Use this fact together with Equations (1) and (2): (3) a = Rα = TR² mg-T I mg m Solve for the tension T: (4) T = 1 + (mR²/I) g Substitute Equation (4) into Equation (2) and solve for a: (5) a = Use a = Ra and Equation (5) to solve for a: a απ = R R + (I/MR) 1 + (I/MR²) g M T m mg Using the results from above, how would you calculate the angular speed of the wheel and the linear speed of the hanging object at t = 2 s, assuming the system is released from rest at t = 0?

A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle as in the figure below. A light cord wrapped around the wheel supports an object of mass m. When the wheel is released, the object accelerates downward, the cord unwraps off the wheel, and the wheel rotates with an angular acceleration. We apply two analysis models here. The object is modeled as a particle under a net force. The wheel is modeled as a rigid object under a net torque. The magnitude of the torque acting on the wheel about its axis of rotation is = TR, where T is the force exerted by the cord on the rim of the wheel. (The gravitational force exerted by the Earth on the wheel and the normal force exerted by the axle on the wheel both pass through the axis of rotation and therefore produce no torque.) Solve for a and substitute the net torque: ΣText = = Ια (1) α = Στοιχε ext I TR I Apply Newton's second law to the motion of the object, Σ Fy = mg - T = ma taking the downward direction to be positive: Solve for the acceleration a: (2) a mg - T = m Equations (1) and (2) have three unknowns: a, a, and T. Because the object and wheel are connected by a cord that does not slip, the translational acceleration of the suspended object is equal to the tangential acceleration of a point on the wheel's rim. Therefore, the angular acceleration a of the wheel and the translational acceleration of the object are related by a = Rα. Use this fact together with Equations (1) and (2): (3) a = Rα = TR² mg-T I mg m Solve for the tension T: (4) T = 1 + (mR²/I) g Substitute Equation (4) into Equation (2) and solve for a: (5) a = Use a = Ra and Equation (5) to solve for a: a απ = R R + (I/MR) 1 + (I/MR²) g M T m mg Using the results from above, how would you calculate the angular speed of the wheel and the linear speed of the hanging object at t = 2 s, assuming the system is released from rest at t = 0?

Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter13: Rotation Ii: A Conservation Approach
Section: Chapter Questions
Problem 16PQ
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A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle as in the figure below. A light cord wrapped around the wheel supports an object of mass m. When the wheel is released, the object accelerates downward, the cord unwraps off the wheel, and the wheel rotates with an angular acceleration. We apply two analysis models here. The object is modeled as a particle under a net force. The wheel is modeled as a rigid object under a net torque. The magnitude of the torque acting on the wheel about its axis of rotation is = TR, where T is the force exerted by the cord on the rim of the wheel. (The gravitational force exerted by the Earth on the wheel and the normal force exerted by the axle on the wheel both pass through the axis of rotation and therefore produce no torque.) Solve for a and substitute the net torque: ΣText = = Ια (1) α = Στοιχε ext I TR I Apply Newton's second law to the motion of the object, Σ Fy = mg - T = ma taking the downward direction to be positive: Solve for the acceleration a: (2) a mg - T = m Equations (1) and (2) have three unknowns: a, a, and T. Because the object and wheel are connected by a cord that does not slip, the translational acceleration of the suspended object is equal to the tangential acceleration of a point on the wheel's rim. Therefore, the angular acceleration a of the wheel and the translational acceleration of the object are related by a = Rα. Use this fact together with Equations (1) and (2): (3) a = Rα = TR² mg-T I mg m Solve for the tension T: (4) T = 1 + (mR²/I) g Substitute Equation (4) into Equation (2) and solve for a: (5) a = Use a = Ra and Equation (5) to solve for a: a απ = R R + (I/MR) 1 + (I/MR²) g M T m mg Using the results from above, how would you calculate the angular speed of the wheel and the linear speed of the hanging object at t = 2 s, assuming the system is released from rest at t = 0?
Transcribed Image Text:A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle as in the figure below. A light cord wrapped around the wheel supports an object of mass m. When the wheel is released, the object accelerates downward, the cord unwraps off the wheel, and the wheel rotates with an angular acceleration. We apply two analysis models here. The object is modeled as a particle under a net force. The wheel is modeled as a rigid object under a net torque. The magnitude of the torque acting on the wheel about its axis of rotation is = TR, where T is the force exerted by the cord on the rim of the wheel. (The gravitational force exerted by the Earth on the wheel and the normal force exerted by the axle on the wheel both pass through the axis of rotation and therefore produce no torque.) Solve for a and substitute the net torque: ΣText = = Ια (1) α = Στοιχε ext I TR I Apply Newton's second law to the motion of the object, Σ Fy = mg - T = ma taking the downward direction to be positive: Solve for the acceleration a: (2) a mg - T = m Equations (1) and (2) have three unknowns: a, a, and T. Because the object and wheel are connected by a cord that does not slip, the translational acceleration of the suspended object is equal to the tangential acceleration of a point on the wheel's rim. Therefore, the angular acceleration a of the wheel and the translational acceleration of the object are related by a = Rα. Use this fact together with Equations (1) and (2): (3) a = Rα = TR² mg-T I mg m Solve for the tension T: (4) T = 1 + (mR²/I) g Substitute Equation (4) into Equation (2) and solve for a: (5) a = Use a = Ra and Equation (5) to solve for a: a απ = R R + (I/MR) 1 + (I/MR²) g M T m mg Using the results from above, how would you calculate the angular speed of the wheel and the linear speed of the hanging object at t = 2 s, assuming the system is released from rest at t = 0?
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