Set 8 Problem number 12

Solution

Generalized Solution
Explanation in terms of Figure(s); Extension
Figure

Problem

A mass of 7.7 kilograms is constrained by a massless rod to move in a circle of radius .6 meters. A torque of 10.8 meter Newtons is applied to the system, which is initially at rest.

What force on the 7.7 kilogram mass, directed perpendicular to the rod, is equivalent to this torque?
What acceleration would this force give the mass?
What angular acceleration would this correpond to?

Find the quantity `tau / (mr ^ 2), where `tau is the torque, m the mass and r the radius of the circle.

What is your result and what is its significance?

Solution

If F is the force applied to the object, then since the force is applied at a distance of .6 meters from the point of rotation and perpendicular to the radial line, the resulting torque will be

torque = `tau = ( .6 meters ) * ( F ).

This is equal to the applied torque of 10.8 meter Newtons.

We solve the equation ( .6 meters)(F) = 10.8 meter Newtons for F, obtaining

F = 18 Newtons.

This force applied to a mass of 7.7 kilograms will result in an acceleration of

a = F / m = 18 Newtons/( 7.7 kilograms) = 2.337662 meters/second ^ 2.

On the given circle, each radian corresponds to a distance equal to the radius .6 meters.

Thus each meter corresponds to 1/ .6 radians, and

2.337662 meters/second ^ 2 corresponds to 2.337662 (1/ .6 ) radians/second ^ 2 = 3.896104 radians/second ^ 2.

Finally, `tau / (mr ^ 2) = ( 10.8 meter Newtons)/[( 7.7 kg)( .6 m) ^ 2] = 3.896104 m N / (kg m ^ 2) = 3.896104 m (kg m / s ^ 2) / (kg m ^ 2) = 3.896104 /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) = 3.896104 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