A cord is wrapped around the rim of a solid uniform wheel 0.290 cm in radius and

Question

A cord is wrapped around the rim of a solid uniform wheel 0.290 cm in radius and of mass 8.60 kg . A steady horizontal pull of 36.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center.Compute the angular acceleration of the wheel.Compute the acceleration of the part of the cord that has already been pulled off the wheel.Find the magnitude of the force that the axle exerts on the wheel. Find the direction of the force that the axle exerts on the wheel.

Explanation / Answer

Given that radius of the wheel is r = 0.290 m

mass   m = 8.60 kg

moment of inertia I = 1/2 m r^2

                               = 1/2 *8.60 * ( 0.290 m )^2

                               = 0.3616 kg m^2

    torque = F * r

                   = 36.0 N * 0.290 m

                  = 10.44 Nm

angular acceleration = / I

                                    = 10.44 / 0.3616

                                     = 28.87 rad /s^2

b ) magnitude and direction of force F_y = - mg , F_x = - P

                            F = F_x ^2 +F_y ^2

                                = 36^2 + ( - 8.60 kg * 9.8 )^2

                                   = 91.64 N

the direction = tan^-1 (5.4808)

                         = 79 degree

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