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

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Homework Help Overview

The discussion revolves around calculating the torque exerted by a mouse on the minute hand of a grandfather clock. The mouse's weight and the angle of the minute hand at a specific time are central to the problem, which involves concepts of torque in a rotational system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the formula for torque and the correct angle to use in calculations. There are questions about the angle between the force vector (weight of the mouse) and the radius vector (position of the mouse on the clock hand).

Discussion Status

Several participants have shared their calculations and expressed confusion over the results. There is an ongoing exploration of the correct angle to use in the torque equation, with some guidance provided regarding the relationship between the vectors involved.

Contextual Notes

Participants are navigating the specifics of the problem setup, including the position of the minute hand and the implications of the angle used in the torque calculation. There is mention of the challenges posed by online homework systems and the need for precision in answers.

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


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


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.
 


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


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).
 


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


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?
 


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 \sin\theta = \sin(180 - \theta).)

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.
 

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