Problem: A mass of 2.9 kilograms is constrained by a massless rod to move in a circle of radius .7000 meters. A torque of 1.8 meter Newtons is applied to the system.
What force on the 2.9 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?
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Solution: If F is the force applied to the object, then since the force is applied at a distance of .7000 meters from the point of rotation and perpendicular to the radial line, the resulting torque will be ( .7000 meters)(F). This is equal to the applied torque of 1.8 meter Newtons. We solve the equation
( .7000 meters)(F) = 1.8 meter Newtons
for F, obtaining
F = 2.571 Newtons.
This force applied to a mass of 2.9 kilograms will result in an acceleration of
a = F / m = 2.571 Newtons/( 2.9 kilograms) = .8865 meters/second ^ 2.
On the given circle, each radian corresponds to a distance equal to the radius .7000 meters. Thus each meter corresponds to 1/ .7000 radians, and .8865 meters/second ^ 2 corresponds to .8865 (1/ .7000) radians/second ^ 2 = 1.266 radians/second ^ 2.
Finally, `tau /mr ^ 2 = ( 1.8 meter Newtons)/[( 2.9 kg)( .7000 m) ^ 2] = 1.266 m N /(kg m ^ 2) = 1.266 m (kg m / s ^ 2) /(kg m ^ 2) = 1.266 /s ^ 2. Note that this result is equal to the angular acceleration, except for the radian unit.
Generalized Response: 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 Newton's Second Law, with angular acceleraton `alpha corresponding to acceleration a, torque `tau (which is what causes the acceleration) to force F and the quantity m r^2, which resists angular acceleration, to mass m (which 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.
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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.