A wheel of radius R = 25 cm and mass 4 kg is placed on a ramp of incline theta =

Question

A wheel of radius R = 25 cm and mass 4 kg is placed on a ramp of incline theta = 25degree as shown to the right. Draw an extended free-body diagram for the wheel on the ramp. Find the linear acceleration of the wheel as it accelerates down the ramp. (Note: you may use I_wheel =1/2m R^2 for the wheel's rotational inertia.) What is the angular acceleration of the wheel? What is the total kinetic energy of the wheel when it reaches the bottom of the ramp? Assume the wheel starts from rest and that the vertical drop is 1.5m (Note: you can answer this and the following questions even if you didn't get (b) or (c)!) What is the velocity of the center of mass and translational kinetic energy of the wheel at the bottom of the ramp? What is the angular velocity and the rotational kinetic energy of the wheel at the bottom of the ramp? What is the angular momentum of the wheel at the bottom of the ramp?

Explanation / Answer

b) The wheel will roll as friction F will provide torque. Let the linear acceleration be a. Since it is pure rolling, angular acceleration alpha = a/R

mg sin 25 - F = ma....equation 1

net torque = FR = I alpha = 0.5 mR^2* a/R

F = 0.5 ma......equation 2

adding the two equations

mg sin 25 degree= 1.5 ma

a = g sin 25 degree /1.5 = 2.761 m/s^2

C) angular acceleration = a/R =2.761/0.25 = 11.04 /s^2

d) Total kinetic ene*rgy = change in potential energy = mgh = 4*9.8*1.5 =58.8 J

e) Ratio between Tr/nslational and rotational KE = 0.5 mv^2 /[0.5 Iw^2] = mv^2/[0.5mr^2*w^2] = 2/1

Translational KE = 58.8*2/3 = 39.2 J

speed v = sqrt[2KE/m] = sqrt[2*39.2/4] = 4.427 m/s

f) Rotational KE = 58.8-39.2 =19.6J

angular velocity w= v/R = 17.7 rad/s

g) angular momentum = Iw = 0.5 mR^2 *w = 0.5*4*0.25*0.25*17.7 = 2.2125 kg m^2/s

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