Set 8 Problem number 12

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s); Extension

• Figure

Problem

Four masses, each of 8.2 /4 kilograms are placed on a massless X shaped frame, one mass at the end of each crossbar. The system rotates about the center, and each mass is 1.3 meters from the center. A rotation-producing torque of 19.9 meter Newtons is applied to the system.

• Suppose that each mass was subjected to a force which would supply 1/4 of this torque. What would be the total of all these forces?
• What would then be the acceleration of each mass?
• What would be the angular acceleration of the system?

Since all the masses are at the same distance r from the axis of rotation, the quantity mr ^ 2, where m is the sum of all the masses, is easily calculated.

• What is the ratio `tau /mr ^ 2 for this system?

Solution

If F is the force applied to the object, then since the force is applied at a distance of 1.3 meters from the point of rotation and perpendicular to the radial line, the resulting torque will be

• torque = `tau = ( 1.3 meters ) * ( F ).

This is equal to the applied torque of 19.9 meter Newtons.

• We solve the equation ( 1.3 meters)(F) = 19.9 meter Newtons for F, obtaining

• F = 15.30769 Newtons.

This force applied to a mass of 8.2 kilograms will result in an acceleration of

• a = F / m = 15.30769 Newtons/( 8.2 kilograms) = 1.866792 meters/second ^ 2.

On the given circle, each radian corresponds to a distance equal to the radius 1.3 meters.

• Thus each meter corresponds to 1/ 1.3 radians, and

• 1.866792 meters/second ^ 2 corresponds to 1.866792 (1/ 1.3 ) radians/second ^ 2 = 1.435994 radians/second ^ 2.

Finally, `tau / (mr ^ 2) = ( 19.9 meter Newtons)/[( 8.2 kg)( 1.3 m) ^ 2] = 1.435994 m N / (kg m ^ 2) = 1.435994 m (kg m / s ^ 2) / (kg m ^ 2) = 1.435994 /s ^ 2.

• Note that this result is equal to the angular acceleration, except for the radian unit.
• Since `tau is in Newtons multiplied by meters of radius, one of the meters (the one from kg m/s^2)  in the resulting unit kg m^2 / s^2 is meters of linear motion and the other in meters of radius. 
• When we divide a linear meter by a meter of radius we are effectively dividing a meter of arc by a meter of radius; since arc / radius = angle in radians, the result is a radian.
• We thus see that the   / s^2 unit is really   rad / sec^2, the unit of angular acceleration.

• Thus `tau / (m r^2) = 1.435994 rad / s^2.

We conclude that dividing the net torque `tau by the quantity m r^2 gives us angular acceleration, in rad / s^2.

• We call the quantity m r^2 the moment of inertia.
• We can therefore say that    net torque / moment of inertia = angular acceleration.
• This is analogous to Newton's Second Law, which can be written   net force / mass = acceleration.

• Torque is analogous to force and moment of inertia to mass.

Generalized Solution

The force F applied at a perpendicular to the moment arm at a point a distance r from the axis of rotation will produce a torque `tau = F * r.

• Applied to a mass m, the force will result in an acceleration a = F / m.
• This acceleration is equivalent to the angular acceleration

• angular acceleration = `alpha = a / r = F / (m r).

Since torque `tau = F * r, the force F is F = `tau / r. As a result we have

• `alpha = F / (mr) = (`tau / r) / (m r) = `tau / (m r^2).

This relationship alpha = `tau / (m r^2) is analogous to (in fact equivalent to) Newton's Second Law, with the following correspondences:

• angular acceleraton `alpha corresponds to acceleration a,
• torque `tau (which is what causes the acceleration) corresponds to force F and
• the quantity m r^2, which resists angular acceleration, to mass m (which as we have seen resists acceleration).

The quantity m r^2 is called the 'moment of inertia' of the mass m at distance r from the center of rotation.

Explanation in terms of Figure(s); Extension

The figure below depicts a force F applied perpendicular to the constraining rod at the position of the mass m.

• The resulting torque `tau is related to the resulting angular acceleration `alpha by the relation
• `alpha = `tau / (m r^2) = `tau / I,

where I stands for m r^2 and is called the moment of inertia of the mass m.

Figure

torque_and_angular_acceleration.gif