Two problems involving rotational movement

AI Thread Summary
The discussion revolves around two problems related to rotational movement. In the first problem, the calculation of torque for a rotating disc is confirmed, with a derived torque of approximately -1.22 Nm, while questioning the relevance of the total rotations made. The second problem involves a bob attached to a rope wrapped around a cylinder, where the potential energy is calculated correctly at 192.2 J, but there is confusion regarding the correct approach to find the bob's speed upon hitting the ground. The participants clarify that the kinetic energy of the cylinder should be calculated using its moment of inertia, not just its mass. Overall, the discussions emphasize the importance of correctly applying rotational dynamics equations in both scenarios.
dzogi
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1. First Problem

Homework Statement


A disc with mass of 50kg and radius of 20cm is rotating with a frequency of 480rpm, and after 50 seconds, as a result of the force of friction, it stops. What's the moment (momentum of force, torque) if during the rotation the disc made 200 rotations?

Homework Equations


M=I\epsilon

The Attempt at a Solution


R=0,2m; m=50kg; f=480min^{-1}=8Hz; t=50s; N=200;
w_0=\frac{2\pi}{1/8}=16\pi rad/s
w = 0rad/s
\epsilon=\frac{w-w_o}{t}=\frac{-16\pi}{50}rad/s
I=0.5mR^2=1

M=I\epsilon=-1.21924Nm

Is this correct? I can't see where to number of total rotations (angular distance) fits in, or maybe it's a distractor?

2. Second problem

Homework Statement


А rope is wrapped around a horizontal cylinder with M=17kg; R=0,1m. A bob with m=5kg is attached at the end of the rope, at height of h=4m above ground. The momentum of inertia of the cylinder is calculated with I=\frac{MR^2}{2}.

a) what's the speed of the bob when it hits ground?
b) calculate the total energy of the system.

Homework Equations



The Attempt at a Solution


I've solved b) pretty easily,
E=mgh=192,2J which conforms to the solution in the book (so, the potential energy of the cylinder is ignored).

I've tried solving a) this way
mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}
If the liner velocity of the cylinder is equal to the speed of the bob at any given time, then we can substitute \omega=\frac{v}{R}
\vdots
Is this approach correct? I don't get the same solution with the one given in the book.
 
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dzogi said:


I've tried solving a) this way
mgh=\frac{mv^2}{2}+\frac{M\omega^2}{2}
If the liner velocity of the cylinder is equal to the speed of the bob at any given time, then we can substitute \omega=\frac{v}{R}
\vdots
Is this approach correct? I don't get the same solution with the one given in the book.


Problem 1 seems fine.

Kinetic energy of the cylinder should be \frac{I\omega^2}{2} and not \frac{M\omega^2}{2}, ain't it?
 
Yes, it was a mistype :)
 
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