How is Torque Calculated for a Mouse on a Clock Hand?

  • Thread starter Lance WIlliam
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In summary, a 55g mouse runs out to the end of the 17cm-long minute hand of a grandfather clock when the clock reads 10 minutes past the hour. The torque exerted by the mouse's weight about the rotation axis of the clock hand can be calculated using the equation torque = rFsin(theta), where r is the distance from the rotation axis to the point where the force is applied, F is the magnitude of the force, and theta is the angle between r and F. At 10 minutes past the hour, the angle between r and F is 60 degrees.
  • #1
Lance WIlliam
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A 55g mouse runs out to the end of the 17cm -long minute hand of a grandfather clock when the clock reads 10 minutes past the hour.

What torque does the mouse's weight exert about the rotation axis of the clock hand?
in Nm

Im using the eqn. (torque)=rFsin(theta)
I put

(torque)=(.17m)(.539the mg)sin(150) and I am wrong...I think its the angle or the force...

The mouses weight should be Mg right? since its the only acting force?
And if a clock is at 2 then its 30 degree above the horizontal and thaking the angle perpendicualr to that is 150 degrees...
I d K...
 
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  • #2


Looks right to me. Why do you think it's wrong?
 
  • #3


well my answer is .09 Nm and it says I am wrong...
 
  • #4


Lance WIlliam said:
well my answer is .09 Nm and it says I am wrong...
Redo your calculation. The formula you had before looked OK to me, but that answer does not.
 
  • #5


I got 4.58e-2...still wrong..I don't see what I am doing wrong?
 
  • #6


Lance WIlliam said:
I got 4.58e-2...still wrong..I don't see what I am doing wrong?
That looks OK to me. Do they want you to include direction? (Like clockwise or counterclockwise?) Those online systems can be picky (and sometimes wrong).
 
  • #7


NOpe they just want it in N m...I hate online homework...thanks for your help.
 
  • #8


Hi Lance WIlliam,

Lance WIlliam said:
A 55g mouse runs out to the end of the 17cm -long minute hand of a grandfather clock when the clock reads 10 minutes past the hour.

What torque does the mouse's weight exert about the rotation axis of the clock hand?
in Nm

Im using the eqn. (torque)=rFsin(theta)
I put

(torque)=(.17m)(.539the mg)sin(150) and I am wrong...I think its the angle or the force...

The mouses weight should be Mg right? since its the only acting force?
And if a clock is at 2 then its 30 degree above the horizontal and thaking the angle perpendicualr to that is 150 degrees...
I d K...


The minute hand is 30 degrees above the horizontal, so that's right. However, I don't believe the sin(150 degrees) in your equation is correct. Do you see what it needs to be?
 
  • #9


alphysicist said:
The minute hand is 30 degrees above the horizontal, so that's right. However, I don't believe the sin(150 degrees) in your equation is correct. Do you see what it needs to be?
At 10 minutes past the hour, the minute hand will be 30 degrees to the right of the vertical or 60 degrees above the horizontal (making the angle between "r" and "F" 150 degrees). Or did I misread the problem?
 
  • #10


if its 10past past the hour its at 2 on the clock so since 3 is a horizontal on the clock the minute hand its 30degrees above it.In toque the (theta) used is the angle between the r vector and the F vector thus they are perpendicular to each other...I get 150 degrees...maybe I am doing it wrong...
 
  • #11


Hi Doc Al,

Doc Al said:
At 10 minutes past the hour, the minute hand will be 30 degrees to the right of the vertical or 60 degrees above the horizontal (making the angle between "r" and "F" 150 degrees). Or did I misread the problem?

At 10 minutes past the hour, the minute hand has made 1/6 of a revolution, so I believe it's 60 degrees from the vertical.
 
  • #12


alphysicist said:
At 10 minutes past the hour, the minute hand has made 1/6 of a revolution, so I believe it's 60 degrees from the vertical.
D'oh! You're right, of course. I was thinking of the minute hand pointing to 1 o'clock, which is 5 minutes past the hour. :redface:

(Good thing I use a digital watch!)
 
  • #13


Lance WIlliam said:
if its 10past past the hour its at 2 on the clock so since 3 is a horizontal on the clock the minute hand its 30degrees above it.In toque the (theta) used is the angle between the r vector and the F vector thus they are perpendicular to each other...I get 150 degrees...maybe I am doing it wrong...
Yes, the r vector is 30 degrees above the horizontal, but the F vector (the weight of the mouse) points down and is thus 90 degrees below the horizontal. They are not perpendicular. (And if they were, that would be 90 degrees, not 150!)

Given that, what's the angle between r and F?
 
  • #14


I get 60 degree as my angle between the two vectors
 
  • #15


Lance WIlliam said:
I get 60 degree as my angle between the two vectors
That will work fine, for your purpose. (Since [itex]\sin\theta = \sin(180 - \theta)[/itex].)

But I'd say to draw yourself a picture. The r vector is 30 degrees above the horizontal, while F is 90 degrees below. What would the angle be between those two vectors?
 
  • #16


THANKYOU for your Help! :)
 
  • #17


You are most welcome!

Thanks to alphysicist for catching the error.
 

What is torque and how is it measured?

Torque is a measure of the force that causes an object to rotate around an axis. It is measured in units of newton-meters (N*m) or foot-pounds (ft-lb) and can be calculated using the equation T = F * r, where T is torque, F is force, and r is the distance from the axis of rotation to the point where the force is applied.

What are some real-life examples of torque?

Some examples of torque in everyday life include twisting a doorknob to open a door, using a wrench to tighten a bolt, and pedaling a bike. Torque is also important in machines such as car engines and industrial equipment.

How does torque affect rotational motion?

Torque is directly proportional to the angular acceleration of an object. This means that the greater the torque, the faster an object will rotate. Torque also determines the direction of rotation, with clockwise torque causing clockwise rotation and counterclockwise torque causing counterclockwise rotation.

What factors affect torque?

The two main factors that affect torque are the force applied and the distance from the axis of rotation. Increasing either of these will result in an increase in torque. Additionally, the angle at which the force is applied can also affect torque, with a perpendicular force resulting in the greatest torque.

How is torque related to power?

Power is the rate at which work is done, and work is equal to force multiplied by distance. Since torque is a measure of force and distance, it is directly related to power. This means that a greater torque will result in a greater power output, which is why high-torque machines are able to do more work in a shorter amount of time.

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