Question 651: Centre of Mass and Rotational Motion

Question

A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at  a distance d from each  other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

Accepted Answer

Given situation is shown in figure.
N
1
= Normal reaction on A
N
2
= Normal reaction on B
W = Weight of the rod
In vertical equilibrium,
N
1
+ N
2
= W Ã¢ÂÂ¦(i)
Torque balance about centre of mass of the rod,
N
1
x = N
2
(d Ã¢ÂÂ x)
Putting value of N
2
from equation (i)
N
1
x = (W Ã¢ÂÂ N
1
)(d Ã¢ÂÂ x)
$$ \Rightarrow $$ N
1
x = Wd Ã¢ÂÂ Wx Ã¢ÂÂ N
1
d + N
1
x
$$ \Rightarrow $$ N
1
d = W(d Ã¢ÂÂ x)
$$ 	herefore $$ $${N_1} = {{W\left( {d - x} ight)} \over d}$$

Answer

$${{W\left( {d - x} \right)} \over x}$$

Answer

$${{W\left( {d - x} \right)} \over d}$$

Answer

$${{Wx} \over d}$$

Answer

$${{Wd} \over x}$$

Centre of Mass and Rotational Motion

A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

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