A slender rod of mass m and length L is pivoted about a horizontal axis through one end and released from rest at an angle of 30degress
above the horizontal. The force exerted by the pivot on the rod at the instant when the rod passes through a horizontal position is :
pls help
Answers
Answer:
ANSWER
BytakingtorqueequationaboutA
τ=
F
×
r
;
τ=Iα
Mgsinθ×l/2=Ml
2
/3×α
α=3gsinθ/2l
Explanation:
ANSWER
The angular velocity of the rod about the pivot when it passes through the horizontal position is given by
mg×
2
L
sin30
o
=
3
mL
2
×
2
ω
2
ω=
2L
3g
Radial acceleration of the centre of mass (as centre of mass is moving in a circle of radius L/2) is given by
a
r
=ω
2
2
L
=
4
3g
Torque about pivot, in the horizontal position, is τ=mgL/2=Iα
α=
mL
2
/3
mgL/2
=
2
3g
Tangential acceleration of the centre of mass, a
t
=
2
L
α=
4
3g
Draw the FBD of the rod an instant when it passes through the horizontal position. Use Newton's second law of equation.
R
1
=ma
r
=
4
3mg
mg−R
2
=m×a
t
=
4
3mg
R
2
=
4
mg
So, reaction force by the pivot on the rod, R=R
1
3
+R
2
2
=
10
mg/4 at an angle of tan
−1
(R
2
/R
1
)[=tan
−1
(1/3)] with the horizontal.