Find frictional torque acting on the pulley?

  • Thread starter Thread starter saturn67
  • Start date Start date
  • Tags Tags
    Pulley Torque
Click For Summary
SUMMARY

The discussion focuses on calculating the average frictional torque acting on a pulley with a 3.75 kg mass and a 2.70 kg mass. The pulley is a uniform solid cylinder with a radius of 4.8 cm and a mass of 0.75 kg. The correct approach involves using the torque equation T1*R - T2*R + frictional torque = I*alpha, where I is the moment of inertia calculated as I = (1/2)mR^2. The final calculated frictional torque is 0.5 mN, confirming the solution's accuracy.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and torque equations
  • Knowledge of moment of inertia calculations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the concept of moment of inertia in detail, particularly for solid cylinders
  • Learn how to derive and apply the equations of motion for connected systems
  • Explore the relationship between linear acceleration and angular acceleration
  • Investigate the effects of friction in rotational systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to torque and rotational motion.

saturn67
Messages
54
Reaction score
0

Homework Statement



A string passing over a pulley has a 3.75 kg mass hanging from one end and a 2.70 kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.8 cm and mass 0.75 kg.

In fact, it is found that if the heavier mass is given a downward speed of 0.23 m/s, it comes to rest in 6.7 s. What is the average frictional torque acting on the pulley?


Homework Equations



sum of torque = I*alpha
I = (1/2)mR^2
a=R*alpha

The Attempt at a Solution



T1*R-T2R- frictional torque=I*alpha
i use that and i got around 0.415mN which is wrong ...

anyone give me some pointer please
 
Physics news on Phys.org
saturn67 said:
T1*R-T2R- frictional torque=I*alpha
That's the torque equation--good! But you'll need other equations to solve for the tensions. (Analyze the forces on each mass.)
 
T1=M1*g-M1*a

T2=M2*a+M2*g

this is right?

do i use the same "a" for each one?
 
saturn67 said:
T1=M1*g-M1*a

T2=M2*a+M2*g

this is right?
Good.

do i use the same "a" for each one?
Yes--they are connected by a string so they must have the same acceleration.
 
... i got 2.95873mN for friction ... still wrong

do we use time for some thing?
 
T1=35.8875N
T1*R=1.7226

T2=-25.839
T2*R=-1.24027

I=0.000864
alpha=4.79167

i use all that and got 2.95873mN
 
saturn67 said:
do we use time for some thing?
Sure. You need it to compute the acceleration. (Note that the acceleration of the heavier mass is upward, so we need to double check the signs in your equations.)
 
T1-M1*g=M1*a

how we use time to find acceleration??

since we have the giving acceleration, time: this is where i have no idea >.<
 
clarification

saturn67 said:
T1=M1*g-M1*a

T2=M2*a+M2*g
Assuming T1 is the heavier mass, let's try it again. Write the equations this way:
T1 - M1*g = + M1*a
T2 - M2*g = - M2*a

Now "a" is just the magnitude of the acceleration, which you can calculate directly.
 
  • #10
saturn67 said:
T1-M1*g=M1*a
Good.

how we use time to find acceleration??
What's the definition of acceleration?
 
  • #11
saturn67 said:
T1*R-T2R- frictional torque=I*alpha
We need to fix the sign of the acceleration in this one also:

-T1*R + T2*R + frictional torque = I*alpha

Now alpha is a positive quantity.
 
  • #12
where can i get T1 or T2

since we have 2 equation and 3 unknow
T1,T2 and a
 
  • #13
You have 3 equations and 3 unknowns. The acceleration is not an unknown. (It's easy to calculate from the given data.)
 
  • #14
a= R* alpha
alpha=a/R??

acceleration = change in v / change in time
 
  • #15
Vf=Vo+at
a=0.034328m/s^2
right?
 
  • #16
saturn67 said:
Vf=Vo+at
a=0.034328m/s^2
right?
Right!
 
  • #17
T1 - M1*g = + M1*a (found T1)
T2 - M2*g = - M2*a (found T2)

I=0.000864

T1*R + T2*R + frictional torque = I*alpha

frictional =0.5mN
 
  • #18
WOOT it work

BIG thank! Doc Al for help me
 
  • #19
You are most welcome. :smile:
 

Similar threads

Replies
1
Views
1K
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K