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Rotational Motion
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Section:
Physics
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© examsnet.com
Question : 16
Total: 54
A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60° with vertical ?
[g is the acceleration due to gravity]
[JEE Adv 2019 P2]
The normal reaction force from the floor on the rod will be
M
g
16
The angular acceleration of the rod will be
2
g
L
The angular speed of the rod will be
√
‌
‌
3
g
2
L
The radial acceleration of the rod’s center of mass will be
3
g
4
Validate
Solution:
→
m
g
(
‌
‌
l
2
‌
sin
‌
60
)
=
‌
‌
m
l
2
3
α
‌
‌
√
3
g
4
=
‌
‌
l
3
α
⇒
α
=
‌
‌
3
√
3
g
4
l
⇒
m
g
‌
‌
l
2
(
1
−
cos
‌
60
)
=
‌
‌
1
2
‌
‌
m
l
2
3
ω
2
‌
‌
g
4
=
‌
‌
l
6
ω
2
ω
2
=
‌
‌
3
g
2
l
→ radial acceleration
a
=
1
2
ω
2
a
=
‌
‌
3
g
4
→
M
g
−
N
=
M
(
‌
‌
3
g
4
×
‌
‌
1
2
+
‌
‌
3
√
3
g
4
l
×
‌
‌
√
3
2
l
)
M
g
−
N
=
M
g
(
‌
‌
3
8
+
‌
‌
9
16
)
N
=
M
g
−
‌
‌
15
M
g
16
=
‌
‌
M
g
16
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