Find angular acceleration of a yoyo

Click For Summary
SUMMARY

The discussion focuses on calculating the angular acceleration of a Yo-Yo consisting of two solid disks with radius R and mass M, and a massless axle with radius r. The key equation used is the sum of torques, represented as Στ = Iα, where I is the moment of inertia (I = 1/2MR²). The final derived formula for angular acceleration (α) is α = 4gr/R², indicating a direct relationship between the axle radius and gravitational acceleration. The confusion regarding the relationship between linear and angular acceleration is clarified, emphasizing the necessity of tension for torque generation.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with moment of inertia calculations
  • Knowledge of linear and angular acceleration relationships
  • Basic principles of gravitational force
NEXT STEPS
  • Study the relationship between linear acceleration and angular acceleration in rotational systems
  • Explore advanced topics in rotational dynamics, such as the effects of varying mass distributions
  • Learn about the implications of different axle radii on torque and angular acceleration
  • Investigate practical applications of angular acceleration in mechanical systems
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding the dynamics of rotating objects, particularly in the context of torque and angular acceleration calculations.

Raziel2701
Messages
128
Reaction score
0

Homework Statement


A Yo-Yo is made up of twp solid disks each with radius R and mass M. The axle has radius r and is essentially massless. The Yo Yo is released to unwind.
Find the angular acceleration of the YoYO as it falls in terms of r, R and g.


Homework Equations


[tex]\Sigma\tau=I\alpha[/tex]

I = 1/2MR^2

I set up the sum of torques to be the tension from the string times r, plus 2MgR, though I don't see what difference there would be if it were 2Mgr:

[tex]\Sigma\tau=Tr +2MgR=I\alpha[/tex]

I setup the following too:

[tex]\Sigma\F_y = T -2Mg = 2Ma[/tex] To solve for T and substitute into my equation for the sum of torques.

Then solving for [tex]\alpha[/tex] I finally get:

[tex]\frac{g(r+R) + ar}{R^2}[/tex] I don't think my answer should have a term for linear acceleration, unless it's gravity too?

I feel like I'm doing things blindly. I know that the sum of torques should equal the moment of inertia times the angular acceleration, but this business with the different radii is confusing me and I am uncertain as to whether or not what I'm doing makes sense.

UNRELATED: I tried previewing the post after doing some modifications and it shows all out of formatting. Please excuse any oddities.
 
Physics news on Phys.org
I think I read too much into the problem. If the yo yo is falling, given the context of the problem there is no mention or too much emphasis on the string being unwounded, so I know that it's rotating but there's nothing to lead me to believe that there's a tension.

So setting up the sum of torques I get:

[tex]\Sigma\tau=2Mgr=I\alpha[/tex]

Solving for alpha I get

4gr/R^2

Still, does this result make sense? I don't really have a way of checking my answer since this homework is just something we turn in. So, I was wondering, if r, the axle was very small, say even zero, then there would be no angular acceleration for the yoyo wouldn't rotate right?

So at least the equation makes sense that way I suppose?
 
If there was no tension there would be no torque and no angular acceleration...you had the right idea above, you just need the relationship between linear acceleration and angular acceleration.
 

Similar threads

Replies
15
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
11
Views
1K
Replies
2
Views
914
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 97 ·
4
Replies
97
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K