Engineering Mechanics

(^646) „„„„„ A Textbook of Engineering Mechanics
Now equating the couple (responsible for rolling) and torque on the body,
F × r=Iα
F × r^2 =Iαr ...(Multiplying both sides by r)
=Ia ...(Q a = rα)
∴ F=
2
22
Ia M k a
rr
= ...(Q I = mk^2 )
Now substituting the value of F in equation (iv),
2

• 2

Mk a

P

r

=Ma

and now substituting the value of P from equation (iii) in the above equation,

2

–2 – 2

Mk a

mg ma

r

=Ma

or mg=

2

(^22)
Mk a
Ma ma
r
++ =
2
(^22)
Mk
aM m
r
⎛⎞
⎜⎟⎜⎟++
⎝⎠
∴ a= 2
(^22)
mg
Mk
Mm
r
⎛ ⎞
⎜ ⎟
⎜⎜ ++ ⎟⎟
⎜⎝ ⎟⎠
The above expression shows that the acceleration of the rolling body is independent of the
value of coefficient of friction.
Example 31.15. Find the acceleration (a) on a solid right circular roller A of mass 20 kg
when it is being pulled by another body B of mass 10 kg along a horizontal plane as shown in
Fig. 31.15.
Fig. 31.16.
The mass B is attached to the end of a string wound round the circumference of the roller.
Assume that there is no slipping of the roller and the string is inextensible.
Solution. Given: Mass of roller (M) = 20 kg and mass of hanging body B (m) = 10 kg
We know that for a solid right circular roller of radius r, 3 k^2 = 0.5 r^2
and acceleration of the roller on the horizontal plane,
a= 2
(^22)
mg
Mk
Mm
r
++^
= 2
2
10 9.8
20 0.5
20 (2 10)
r
r
×
⎛⎞×
⎜⎟⎜⎟+× +
⎝⎠
m/s^2
= 1.96 m/s^2 Ans.