Solve Kinetic Questions - 2 Examples

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Homework Help Overview

The discussion revolves around two physics problems: one involving the calculation of muscle mass percentage during a chin-up and the other concerning the kinetic energy of a rotating bar. Both problems are situated within the context of mechanics, specifically focusing on work and energy concepts.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to apply equations related to work and kinetic energy but express confusion regarding their calculations and the concepts involved. Some participants question their mathematical steps and seek clarification on the definitions and applications of rotational kinetic energy.

Discussion Status

The discussion is ongoing, with participants actively seeking help and clarification on their approaches. Some guidance has been offered regarding the need to consider rotational kinetic energy and the proper setup for the chin-up problem, but no consensus has been reached on the correct solutions.

Contextual Notes

Participants note that they are at different stages in their coursework, with one mentioning they have not yet covered rotational kinetic energy. There is also a recurring theme of uncertainty regarding the calculations and assumptions made in the problems.

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Homework Statement


While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)

Homework Equations


W = F * s

The Attempt at a Solution


m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg

Then I take this and divided by total mass 70kg to get the percentage but I got the wrong answer. I don't know why...what am I doing wrong here?

Homework Statement


Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar?

Homework Equations


KE = 1/2m (vf^2 - vo^2)

The Attempt at a Solution


I didn't know how to get this started but I gave it a shot. First, I took the 5 revolution per 3 seconds and converted them to velocity.

5rev/3s (2*pie*r/rev) = 20.9 m/s

Then I take that answer and plug it into the kinetic equation but as expected, I was wrong. This kinetic stuff is really confusing or is it just me? :confused: Pls help?
 
Last edited:
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Peach said:

Homework Statement


Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds.

What is the problem question here?
 
Oops. :redface: I totally forgot.

Question: What is the kinetic energy of this bar?
 
Peach said:

Homework Statement


While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)


Homework Equations


W = F * s

The Attempt at a Solution


m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg

Check your math. Your equation seems right, but I get a different answer when I plug in the values.
 
Peach said:
Oops. :redface: I totally forgot.

Question: What is the kinetic energy of this bar?

Excellent. Now, how can you calculate the angular velocity of the bar from the given information? The units should be [rad/s].

Further on, what is the definition of rotational kinetic energy?
 
Peach said:

Homework Statement


Rotating Bar. A thin, uniform 12.0-kg bar that is 2.00 m long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar?

Homework Equations


KE = 1/2m (vf^2 - vo^2)

The Attempt at a Solution


I didn't know how to get this started but I gave it a shot. First, I took the 5 revolution per 3 seconds and converted them to velocity.

5rev/3s (2*pie*r/rev) = 20.9 m/s

Then I take that answer and plug it into the kinetic equation but as expected, I was wrong. This kinetic stuff is really confusing or is it just me? :confused: Pls help?

First of all, your radius of the bar is 1m. Don't get confused.

Second of all, look into Rotational Kinetic Energy. You're using the linear kinetic energy equation
 
Last edited by a moderator:
1) I checked my math already and I still get 17.86kg. It should be 70/3.92 right? Are you sure I'm not missing anything?

2) I haven't learned about rotational kinetic energy yet. That's on chapter 9, I'm only on chapter 6.

Edit: Is the angular speed (5rev/3s)(2pie/rev) = (10pie*rad/3s)?
 
Last edited:
Can someone pls pls pls help me with the first problem? I can't figure it out at all, I don't know where I went wrong. I've been punching numbers for the last hour and I still can't seem to get the right answer and this, I know, is a really easy problem and yet I'm going crazy over it. Sigh.

1. Homework Statement
While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)


2. Homework Equations
W = F * s

3. The Attempt at a Solution
m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg
 
First problem

Peach said:

Homework Statement


While doing a chin-up, a man lifts his body 0.40 m. The muscles involved in doing a chin-up can generate about 70 J of work per kilogram of muscle mass. If the man can just barely do a 0.40-m chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical 70-kg man with 14% body fat is about 43%.)


Homework Equations


W = F * s

The Attempt at a Solution


m = w/(g*s)
m = 70J/(9.8m/s^2 * .40m
m = 17.86 kg

Then I take this and divided by total mass 70kg to get the percentage but I got the wrong answer. I don't know why...what am I doing wrong here?
Not quite sure what you are doing here. Try this. Call the man's total mass M and his muscle mass m. Figure out: The amount of work required to do one chin up; the amount of work his muscle mass can generate. Set them equal and solve for m/M.
 

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