Chapter 31 : Kinetics of Motion of Rotation „„„„„ 633

First of all, consider the motion of the body, which is coming down. We know that the forces
acting on it are m.g. (downwards) and P (upwards). As the body is moving downwards, therefore
resultant force acting on it

=mg – P ...(i)
Since the body is moving downwards, with an acceleration (a), therefore force acting on it.
=ma ...(ii)
Equating equations (i) and (ii),
mg – P=ma ..(iii)
Now consider motion of the pulley, which is rotating about its axis due to downward motion
of the body tied to the string. We know that linear acceleration of the body is equal to the angular
acceleration of the pulley.

∴ a=ra ...(iv)

and torque, T= Tension in the string × Radius of the pulley

=P × r ...(v)

We also know that torque on the pulley,

T=Iα ...(vi)

Equating equations (v) and (vi),

P × r=Iα

Pr^2 =Iαr ...(Multiplying both sides by r)

=Ia ...(Q a = rα)

∴ P= 2

Ia

r

...(vii)

Substituting the value of P in equation (iii),

• 2
  Ia
  mg
  r

=ma

2

Ia

ma

r

+ =mg

2

I

am

r

⎛⎞

⎜⎟+

⎝⎠

=mg

∴ a= 2

(^22)
mg mg
I Mk
m m
r r

⎛⎞⎛ ⎞+
⎜⎟⎜⎟⎜⎟+
⎝⎠⎝⎠
...(viii)
Now substituting the value of a in equation (vii),
P= 22
2
Img Img
rmrII
m
r
×=
⎛⎞+ +
⎜⎟
⎝⎠
...(ix)
Note: If the pulley is a solid disc, then its mass moment of inertia (I) is 0.5 Mr^2. Now
substituting the value of I in equation (viii) and (ix).
a= 2
2
2
0.5 0.5^2
mg mg mg
Mr mMmM
m
r

++

• 
  

  and P= 2
  Img
  mr +I^

  
  

  2
  22
  0.5
  0.5^2
  Mrmg Mmg
  mr Mr mM

  
  


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