Acceleration of a pulley/ rotational kinematics - pleaaaase help

Click For Summary
SUMMARY

The discussion focuses on calculating the angular acceleration (α) of a uniform solid cylinder when a block attached to a string wrapped around it descends. The equations of motion are established using Newton's second law (F=ma) and the torque equation (τ=Iα), where I is the moment of inertia for a cylinder (I=0.5mr²). The solution involves substituting the linear acceleration (a) into the torque equation and solving for α, ultimately leading to the conclusion that α can be derived from the relationship between tension (T), mass (m), radius (r), and gravitational acceleration (g).

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of rotational dynamics and torque (τ=Iα)
  • Familiarity with moment of inertia for a uniform solid cylinder (I=0.5mr²)
  • Basic algebraic manipulation and system of equations
NEXT STEPS
  • Study the derivation of angular acceleration in rotational motion problems
  • Learn about the relationship between linear and angular kinematics
  • Explore the concept of tension in systems involving pulleys and blocks
  • Investigate real-world applications of rotational dynamics in engineering
USEFUL FOR

Physics students, mechanical engineers, and anyone studying dynamics and rotational kinematics will benefit from this discussion.

jaded18
Messages
149
Reaction score
0
A string is wrapped around a uniform solid cylinder of radius r, as shown in the figure. The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m.
http://session.masteringphysics.com/problemAsset/1011172/10/MAD_ia_6.jpg
Find the magnitude alpha of the angular acceleration of the cylinder as the block descends.

_________
I know that F=ma = -(T-mg) and
magnitude of torque that acts on pulley = I(alpha) = -Tr
and -T = 1/2 (m*r*alpha) if we substitute (0.5m(r^2)) which is the moment of inertia of a uniform cylinder in the second equation
and I also know that a = -alpha (r)

I guess I just don't see how i can use system of equations to eliminate T and still get an answer with variables r and g in the end ..
 
Physics news on Phys.org
All you need to do is solve your equations...

ma = -(T-mg) (1)

-T = 1/2 (m*r*alpha) (2)

a = -alpha (r) (3)

solve for alpha in (3)... plug it into (2).

and you're left with

ma = -(T-mg)
T = 1/2 ma

solve for a then you can get alpha using (3)... remember the question asks for the magnitude... so once you solve it, don't worry about the minus sign if you get one.
 
thanks ^^
 
Last edited:

Similar threads

Replies
15
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 6 ·
Replies
6
Views
948
Replies
4
Views
862
  • · Replies 4 ·
Replies
4
Views
13K
  • · Replies 78 ·
3
Replies
78
Views
10K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K