Moment of inertia of a tire

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Homework Statement:

Jane drives her sportscar East along the road at a speed of 34 mph. Each tire has an inner radius of 0.4 m, an outer radius of 0.52 m, and a mass of 6 kg.

Jane sees a sign for an elementary school, so she gradually slows down to 19 mph over a period of 5 seconds, decelerating at a constant rate.

What was the torque applied to that tire during the period of acceleration? Provide both a magnitude and direction.

Relevant Equations:

I=(1/2)(M)(radius1^2 + radius2^2)
I=(1/2)(6kg)(0.4^2 + 0.52^2) = 1.29 kg*m^2
initial:
34 mph= 15.2m/s
15.2m/s = (ω) (0.52m)
ω= 29.2 rad/s
after:
19 mph = 8.5m/s
8.5m/s = ω(0.52m)
ω= 16.3 rad/s

acceleration = (16.3 rad/s - 29.2 rad/s) / 5s= -2.58 rad/s^2
torque= |-2.58 rad/s^2 |(1.29 kg*m^2 ) = 3.3 N*m
I am confused on what the direction of the torque is.
 
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Answers and Replies

  • #2
TSny
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Are you familiar with the concept of torque as a vector quantity? If so, then the answer for the direction of the torque would be the direction of the torque vector.

Your solution for the magnitude of the torque looks good, although I have not checked the numerical calculations.
 
  • #3
Delta2
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Torque's direction is usually perpendicular to the plane of rotation. It is not the direction of rotation (clockwise or counterclockwise) if that's what you 've been thinking.
 
  • #4
haruspex
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The direction of a rotation vector is a matter of convention. Standard is the right hand grip:
1572328105072.png

Curl the fingers of your right hand to mimic the rotation. The thumb points in the direction of the corresponding vector.
In the case of a slowing car from the driver's perspective, the top of the tire is moving away (forwards) more slowly, so for the acceleration your hand should be palm up. That makes the vector point to the right.

Btw, note that what you are finding here is the net torque. The brakes will apply a much greater torque, but most of it will be countered by friction from the road.
 
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  • #5
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The torque points to South
 
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  • #6
haruspex
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