What is the correct representation of torque in the provided physics problem?

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In summary, The conversation is about a confusion in a physics problem involving torque and angular acceleration. The solution provided considered the torque positive if it resulted in linear acceleration in the direction of the applied force, leading to opposite results. The person involved did not make any mistake and the solution should not have changed the sign convention.
  • #1
tmiddlet
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Hello pf!
I'm self-studying physics on MIT OCW, and I'm confused about one of the challenge problems. (attached)
I'm looking at problem #2, and I thought I had it understood, but when I looked at the solution the first thing that is stated is that τ = Rf - bF (f = the force of friction). For some reason, I got τ = bF - Rf. I took the cross product of <0,-b> and <F,0> which if bF and the cross product of <0,-R> and <-f,0>, which is -Rf.
I just don't see how we came with the opposite of the answer. Where did I go wrong?
 

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  • #2
You are right about the sign of the torque, but the yo-yo will accelerate to the right so the angular acceleration of rolling is negative. The solution considered the torque positive if it resulted in linear acceleration in the direction of the applied force.

ehild
 
  • #3
To clarify, you didn't do anything wrong. Both angular acceleration and Rf torque are negative, while bF torque is positive. Even though the linear acceleration corresponds to negative angular acceleration, the provided solution should not have changed the sense of the normal sign convention for torque and angular acceleration.
 

FAQ: What is the correct representation of torque in the provided physics problem?

What is the science behind pulling a yo-yo on a table?

Pulling a yo-yo on a table involves the principles of gravity, friction, and rotational motion. When the yo-yo is released, gravity pulls it down towards the table, while the friction between the yo-yo and the table surface creates a force that slows down its descent. As the yo-yo spins, the force of friction also helps to keep it in motion.

Why does a yo-yo return to the hand after being pulled on a table?

A yo-yo is designed with a string attached to a spool, allowing it to spin freely. When the yo-yo reaches the end of the string, the string creates tension on the spool, causing it to rewind back to the hand. This is due to the conservation of angular momentum, which states that an object in motion will continue to move in the same direction unless acted upon by an external force.

What factors affect the yo-yo's performance on a table?

The performance of a yo-yo on a table can be affected by factors such as the weight and shape of the yo-yo, the length and material of the string, the surface of the table, and the force and angle at which it is pulled. These factors can impact the amount of friction and rotational motion, ultimately affecting the yo-yo's speed and duration of spin on the table.

Can a yo-yo be pulled on any surface?

Yes, a yo-yo can be pulled on any surface as long as it has enough friction to slow down its descent. However, the performance of the yo-yo may vary depending on the surface. For example, a smoother surface may result in less friction and a shorter spin duration, while a rougher surface may provide more friction and a longer spin duration.

How does pulling a yo-yo on a table relate to real-world applications?

The principles of gravity, friction, and rotational motion involved in pulling a yo-yo on a table can be applied to various real-world scenarios. For example, understanding these principles can help engineers design better wheels for cars and roller coasters, improve the performance of ball bearings, and even aid in the development of space exploration vehicles.

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