Calculating Tension and Acceleration in a Rotating System

  • Thread starter bard
  • Start date
In summary, to solve for the tension and acceleration of an object tied to a wheel with a light string, you need to apply Newton's 2nd law to both the mass and the wheel, using the rotational form for the wheel. By relating the two equations, you can solve for the tension and acceleration of the object, which are given by T = mR^2g/(I+mr^2) and a = mgr/I+mR^2, respectively.
  • #1
bard
65
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An object of mass m is tied to a light string wound around a wheel that has a moment of inertia I and radius R. The wheel bearing is frictionless and the string does not slip. find the tension and the acceleration of the object.

I think mg-T=ma


and that I=MR^2

thats all i know can someone help me please

thank you
 
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  • #2
You need to apply Newton's 2nd law to each body: the mass and the wheel.
Originally posted by bard
I think mg-T=ma
Right. That's Newton's 2nd law applied to the mass.

Now apply it to the wheel, using the rotational form: Torque = I*α (you tell me the torque).

To relate the two equations, note that a = R*α.
 
  • #3
Hi Doc Al,

Almost solved it
the a i use here will be angular acceleration

so

net torque=T*R or

Ia=T*R

Ia=(mg-mRa)R

Ia=mgr-mR^2a

Ia+mgR^2a=mgr
a(Ia+mR^2)=mgr

a=mgr/I+mR^2

for acceleration the tension is a=mR^2g/(I+mr^2)
 

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