Question
A uniform slender bar AB of mass m is suspended as shown from a small cart of the same mass m. Neglecting the effect of friction, determine the accelerations of points A and B immediately after a horizontal force F has been applied at B
Answer: Option D
:
D
Let the acceleration of bar and cart be a and a' respectively; alsoαbe the angular acceleration of the bar.
Writing newton's equation of motion
For the bar: F - F' =ma ...(i)
For the cart: F' = ma' ...(ii)
Torque equation for rod about center of mass
For rotation of the rod (F + F)L2=ML212α
F+F'=MLα6 ...(iii)
The acceleration of point A on the rod will be same as the acceleration of the cart,
-a' =αL2-aora' =a−L2α →aA=→aAC+→ac ...(iv)
On solving eqation (i), (ii), (iii) and (iv), we get, α=7F5mandα=18F5mL
|a'| =|aA|=a−L2−α−2F5m; |aB|=a+L2α=16F5m
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:
D
Let the acceleration of bar and cart be a and a' respectively; alsoαbe the angular acceleration of the bar.
Writing newton's equation of motion
For the bar: F - F' =ma ...(i)
For the cart: F' = ma' ...(ii)
Torque equation for rod about center of mass
For rotation of the rod (F + F)L2=ML212α
F+F'=MLα6 ...(iii)
The acceleration of point A on the rod will be same as the acceleration of the cart,
-a' =αL2-aora' =a−L2α →aA=→aAC+→ac ...(iv)
On solving eqation (i), (ii), (iii) and (iv), we get, α=7F5mandα=18F5mL
|a'| =|aA|=a−L2−α−2F5m; |aB|=a+L2α=16F5m
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