A Spool On the Move A spool rests at the top of an incline. It is made of two un

Question

A Spool On the Move A spool rests at the top of an incline. It is made of two uniform disks of radius R and mass m connected by a horizontal cylinder of the same mass but a third of the radius (It is shaped similar to a dumbell). A massless string is wound around the cylinder and attached to the top of the incline as shown. When the spool is released it unwinds as it rolls down the incline without slipping. a. Give a simplified symbolic expression for the moment of inertia of the spool about the center of mass in terms of m and R. Now suppose the spool takes 5.5 seconds to makes 3 full revolutions at constant angular acceleration down the incline. Also assume m = 500 g, R = 10 cm, and a = 150 b. What is the angular acceleration of the spool about the center of mass? C. What is the linear speed of the center of mass after the 5.5 seconds? d. By summing torques, find the tension in the string as it rolls down the incline. HINT: In part (a) the moment of inertia about an axis through the center of mass was found. The moment of inertia about an axis parallel to that axis (ie. not through the center of mass) will always be greater than the moment of inertia through the center of mass by an amount equal to mhz where h is the distance between the center of mass axis and the new, parallel axis. This is called the Parallel Axis Theorem. e. What is the magnitude and direction of the (constantfriction?

Explanation / Answer

moment of inertia of cylinder = (1/2)*m*(R/3)^2

moment of inertia of disk = (1/2)*m*(R^2 + (R/3)^2)

moment of inertia of spool about the center I = (1/2)*m*(R^2 + 2R^2/9)

Ic = (11/18)*m*R^2

moment of inertia about the point of contact I = Ic + 2m*R^2

I = 2.61 *mR^2

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displacement theta = 3 revolutions = 6pi rad

time t = 5.5 sec

theta = wo*t + (1/2)*alpha*t^2

6pi = (1/2)*alpha*5.5^2

angular acceleration = alpha = 1.25 rad/s^2

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(c)

linear speed = v = R*w = R*alpha*t = 0.1*1.25*5.5 = 0.687 m/s

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d)

net torque = I*alpha

T*R/3 = (2.61)*m*R^2*alpha

T = (2.61*3)*m*R*alpha

T = (2.61)*0.5*0.1*1.25

T = 0.163 N

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e)

linear acceleration a = R*alpha = 0.1*1.25 = 0.125 m/s^2

T + f - 2mg*sintheta = 2m*a

0.163 + f - (2*0.5*9.8*sin15) = 2*0.5*0.125

frictional force = 2.5 N

direction up the incline

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