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JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions
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© examsnet.com
Question : 24
Total: 100
a string is wound around a hollow cylinder of mass
5
k
g
and radius
0.5
m
. If the string is now pulled with a horizontal force of
40
N
,
and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)
[11 Jan. 2019 II]
20
r
a
d
∕
s
2
16
r
a
d
∕
s
2
12
r
a
d
∕
s
2
10
r
a
d
∕
s
2
Validate
Solution:
From newton's second law
40
+
f
=
m
(
R
α
)
.......(i)
Taking torque about 0 we get
40
×
R
−
f
×
R
=
I
α
40
×
R
−
f
×
R
=
m
R
2
α
40
−
f
=
m
R
α
.......(ii)
Solving equation (i) and (ii)
α
=
40
m
R
=
16
r
a
d
∕
s
2
© examsnet.com
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