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A uniform rod of length l and mass M piv...

A uniform rod of length l and mass M pivoted about its end as shown in Fig. and is free to rotate in the vertical plane about the pivot. The rod is released from rest in the horizontal position.
(a) What is the initial angular acceleration of the rod?

(b) Find the initial acceleration of the right end of the rod?
( c) Find normal contact force due to hinge when rod has rotated through angle `theta` as shown in Fig.

Text Solution

Verified by Experts

(1) It is easy to imagine that rod rotates clockwise around the pivot at the left end, when it is released. The rod will be experiencing two forces, one is the weight, effectively acting on center of mass and the other is the force due to hinge or pivot, acting through the hinge. (2) When we write torque about the pivot the torque due to second force will become zero because its moment arm is zero. Now the torque is due to only the gravitational force on the rod. For the rotation axis considered crossing the pivot, the rod is a rigid body under a net torque.
Calculation: We know the gravitational force acts effectively at the center of mass of the rod as shown in Fig.
The magnitude of the torque due to the weight about an axis through the pivot,
`tau=Mg(l/2)`

Using Eq. to obtain the angular acceleration of the rod,
`alpha=tau/I=(Mg(l/2))/(1/3Ml^(2))=(3g)/(2l)`
(b) Compute: Using Eq. with r = l for the initial acceleration of the right end of the rod, we get
`alpha_(i)=lalpha=3/2g`
( c) Calculations: We know the gravitational force effectively acts at the center of mass of the rod as shown in Fig. The magnitude of the torque due to the weight about an axis through the pivot is given by
`mgcosthetaxxl/2=(ml^(2))/3xxalpha`
`alpha=(3gcostheta)/(2l)`
By energy conservation,
`mg1/2sintheta=1/2(ml^(2))/3omega^(2)`
`omega=sqrt((3gsintheta)/l)`
`a_(r)=omega^(2)l/2=(3gsintheta)/lxxl/2=(3gsintheta)/2`
`a_(t)=alpha1/2=(3gcostheta)/4`
`F_(N_(1))-mgsintheta=m(3gsintheta)/2`
`F_(N_(1))=(5mgsintheta)/2`
`mgcostheta-F_(N_(2))=m(3gcostheta)/2`
`F_(N_(2))=(mgcostheta)/4`
`F_(N)=sqrt(F_(N_(1)^(2))+F_(N_(2)^(2)))`
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