Torque Req. for Fixed Rot. Acc.

[Beyond Aphelion]

I'm having difficulty with this question:

A day-care worker pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 10.1s. Assume the merry-go-round is a disk of radius 2.30m and has a mass of 830kg, and two children (each with a mass of 25.4kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque.

Alright, this is my process; although, I know my end result is wrong:

I used the angular kinematic equation to solve for the angular acceleration.

(ωf = ωi + αΔt)

I got α = 0.186629 rad/s² (approx.) after converting from rpm's.

The equation I have for torque is:

τ = mr²α

But, since we're working with a disk, I = ½MR².

Therefore, I solved for torque using the equation:

τ = ½MR²α

I'm moderately confident with myself at this point, although I realize I can be completely off, but I think I'm screwing up what to use for mass.

I plugged in the mass of the merry-go-round plus the mass of the two children.

M = 880.8 kg

Most likely, this is where my reasoning is flawed. I've just recently been introduced to torque, and it is honestly confusing me.

Anyway. The answer I got:

τ = ½MR²α = τ = ½(880.8 kg)(2.3)²(0.186629 rad/s²) =

434.79 N*m

This is the wrong answer, I know. But it is the best I could come up with based on the information my textbook is giving me. Any advice would be helpful.

 

[nrqed]

I'm having difficulty with this question:

A day-care worker pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 10.1s. Assume the merry-go-round is a disk of radius 2.30m and has a mass of 830kg, and two children (each with a mass of 25.4kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque.

Alright, this is my process; although, I know my end result is wrong:

I used the angular kinematic equation to solve for the angular acceleration.

(ωf = ωi + αΔt)

I got α = 0.186629 rad/s² (approx.) after converting from rpm's.

The equation I have for torque is:

τ = mr²α

But, since we're working with a disk, I = ½MR².

Therefore, I solved for torque using the equation:

τ = ½MR²α

I'm moderately confident with myself at this point, although I realize I can be completely off, but I think I'm screwing up what to use for mass.

For a point mass, I is MR^2 (where R is the distance from the point mass to the axis of rotation). For a disk, the moment of inertia is 1/2 MR^2. What you have to do here is to calculate the total moment of inertia, with is $I_{total}=I_{child#1} + I_{child#2} + I_{disk}$

Use this for the total moment of inertia. barring any algebra mistake, this should work.

Patrick

 

[kyle8921]

As a scientist, it is important to approach problems with a critical and analytical mindset. In this situation, it seems that your approach is on the right track, but there are a few areas where some adjustments could be made.

Firstly, when calculating torque, it is important to consider the distance from the axis of rotation to the point where the force is applied (in this case, the hand pushing the merry-go-round). This distance is known as the lever arm and is denoted by "r" in the equation τ = rFsinθ. In this problem, the force is applied tangentially, so the angle θ would be 90 degrees, making sinθ = 1. Therefore, the equation for torque in this situation would be τ = rF.

Secondly, the mass used in the equation for torque should only include the mass of the merry-go-round, as it is the only object that is rotating. The children's masses should not be included in this calculation.

Finally, it is important to convert all units to a consistent system. In this problem, the given radius is in meters, but the mass is given in kilograms. It would be helpful to convert the mass to grams (1 kg = 1000 g) so that the units for torque are in Newton-meters (N*m).

With these adjustments, the correct calculation for torque would be:

τ = rF = (2.3 m)(880.8 kg)(0.186629 rad/s²) = 387.3 N*m

I hope this helps clarify any confusion and assists in your understanding of torque. Remember to always carefully consider the variables and units involved in a problem and approach it with a systematic approach. Good luck!