How Much Force is Needed to Stop a Spinning Grinding Wheel in 20 Seconds?

  • Thread starter tandoorichicken
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In summary, by plugging in the given values and using the equations for rotational motion, we can calculate the angular acceleration of the wheel and the frictional force acting on it. From there, we can use the formula for torque to find the normal force needed to stop the wheel in 20 seconds. The force needed to press against the wheel would be equal to the normal force multiplied by the coefficient of friction.
  • #1
tandoorichicken
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A piece of metal is pressed against the rim of a 1.6kg, 19-cm-diameter grinding wheel that is turning at 2400 rev/min. The metal has a coeffiecient of friction of .85 with respect to the wheel. When the motor is cut off, with how much force must you press to stop the wheel in 20 sec.?

Not sure I did this right, here's what I did:
2400 rev/min = [tex] 80 \pi[/tex] rad/sec. = [tex] \omega_0 [/tex].
t=20, and [tex] \omega_f = \omega_0 + \alpha t [/tex]. After plugging everything in (w_f=0) I got [tex] \alpha = -40 \pi [/tex] rad/sec^2. Since [tex] a = r \alpha [/tex], I got a = [tex] -3.4 \pi [/tex] m/sec^2. Then somehow from there I jumped to an answer of [tex] 4.624 \pi [/tex] N. I don't think that's right.
 
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  • #2
Originally posted by tandoorichicken
Not sure I did this right, here's what I did:
2400 rev/min = [tex] 80 \pi[/tex] rad/sec. = [tex] \omega_0 [/tex].
t=20, and [tex] \omega_f = \omega_0 + \alpha t [/tex]. After plugging everything in (w_f=0) I got [tex] \alpha = -40 \pi [/tex] rad/sec^2.
[itex]\alpha = -4\pi[/itex]. Now that you've calculated [itex]\alpha[/itex], how do you relate this to force? Make use of [itex]\tau = I\alpha[/itex] to find the torque and thus the frictional force. You will need to know the rotational inertia (I) of a disk; look it up.
 
  • #3
So once I find the linear force of the wheel do I just multiply that by the coefficient of friction to find the force that I push with?
 
  • #4
Originally posted by tandoorichicken
So once I find the linear force of the wheel do I just multiply that by the coefficient of friction to find the force that I push with?
No. The force you press with is the normal force; the force producing the torque on the wheel is the friction. Ffriction=μN.
 

FAQ: How Much Force is Needed to Stop a Spinning Grinding Wheel in 20 Seconds?

What is torque and how is it calculated?

Torque is a measure of the rotational force applied to an object. It is calculated by multiplying the force applied to an object by the distance from the pivot point to the point where the force is applied.

How is torque different from linear force?

Torque is a rotational force, while linear force is a straight, pushing or pulling force. Torque involves an object rotating around a fixed point, while linear force involves an object moving in a straight line.

What is angular acceleration and how is it related to torque?

Angular acceleration is the rate of change of an object's angular velocity, or how quickly it is rotating. It is related to torque through Newton's Second Law for rotational motion, which states that the net torque on an object is equal to the object's moment of inertia times its angular acceleration.

How does torque affect the motion of an object?

Torque causes an object to rotate around an axis. The direction of the torque determines the direction of the rotation, while the magnitude of the torque determines the speed of the rotation. A larger torque will cause the object to rotate faster, while a smaller torque will cause it to rotate slower.

What are some real-life examples of torque and angular acceleration?

Some examples of torque and angular acceleration in everyday life include opening a door, using a wrench to tighten or loosen a bolt, and riding a bicycle. In all of these cases, a force is applied at a distance from the pivot point, causing the object to rotate. Other examples include the rotation of a carousel, the movement of a spinning top, and the rotation of a planet around its axis.

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