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Mass suspended from a pulley

  • #1
4
0

Homework Statement


A mass of 0.5kg is suspended from a flywheel, if the mass is released from rest and falls a distance of 0.5m in 1.5s calculate:

A, the Linear acceleration of the mass
B, the angular acceleration of the wheel
C, the tension in the rope
D, the friction torque.

Known information
Mass of wheel 3 kg Mp
Outside radius of wheel 0.3m
Radius of gyration 0.212 m

Homework Equations




The Attempt at a Solution



A,
mg - T = ma
TR=Ia/R
mg-(Ia/R^2) = ma
a= mg/(m+(I/R^2))
Swapping I out for I=(MpR^2)/2
gives
a=mg/(m+((MpR^2)/R^2)
a=mg/(m+(Mp/2)
a=2.4525
2.45 m/sec^-2

B,
Angular acceleration = a/R
2.4525/0.3 = 8.175
8.175 rads/sec^-2

C,
T=mg-ma
T=0.5*9.81-0.5*2.4525
T=3.68
3.68N

D, - The problem!
Torque (TNET) = acceleration torque (TA) + frictional torque (TF)
TF = TR
3.68*0.3
1.103625

TA = Angular acceleration*I
I = MpR^2/2
I = 3*0.3^2/2
I = 0.135
0.135*8.175
= 1.103625

Thus suggestion there is no frictional Torque,

I have spent a few weeks on an off with this question, I have looked at out threads on here, the lesson provided with the module and YOUTUBE. I am still not able to work out were I am going wrong, hope someone can help, Thanks,
 

Answers and Replies

  • #2
4
0
Funny how posting that that prompted thought,

Should net torque use the radius of gyration instead of radius, thus giving a TNET of 0.779895, apply the same to the TA calculation it gives 0.551258
thus
TF = TNET - TA
TF = 0.228637

Or have I gone wrong further up the chain?
 
  • #3
gneill
Mentor
20,792
2,770
Hi SWINEFLU, Welcome to PF.

You have a problem with part A. You're looking for the linear acceleration of the mass and you're given distance and time, so why not avoid all the rotational motion involvement to begin with? Simple kinematics will yield the linear acceleration.
 
  • #4
gneill
Mentor
20,792
2,770
Funny how posting that that prompted thought,

Should net torque use the radius of gyration instead of radius, thus giving a TNET of 0.779895, apply the same to the TA calculation it gives 0.551258
thus
TF = TNET - TA
TF = 0.228637

Or have I gone wrong further up the chain?
The radius of gyration will let you compute the moment of inertia using the given mass of the wheel. What's the relationship between radius of gyration, moment of inertia, and mass? This should be one of your Relevant Equations.
 
  • #5
4
0
Hi SWINEFLU, Welcome to PF.

You have a problem with part A. You're looking for the linear acceleration of the mass and you're given distance and time, so why not avoid all the rotational motion involvement to begin with? Simple kinematics will yield the linear acceleration.
Thanks gneill, I understand were you are coming from as this is what I have read online in several places, but surely the rotating of the mass of which the string is attached to affects the overall acceleration of the mass?
 
  • #6
gneill
Mentor
20,792
2,770
Thanks gneill, I understand were you are coming from as this is what I have read online in several places, but surely the rotating of the mass of which the string is attached to affects the overall acceleration of the mass?
It does, yes, but whatever affect it has results in the observed motion which is that the mass travels a given distance in a given time, hence a deducible acceleration (assuming linear motion and constant acceleration).
 

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