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Figure 4.25 shows a child's toy, which has the shape of a
Chapter 4, Problem 4.30(choose chapter or problem)
Figure 4.25 shows a child's toy, which has the shape of a cylinder mounted on top of a hemisphere. The radius of the hemisphere is R and the CM of the whole toy is at a height h above the floor.
(a) Write down the gravitational potential energy when the toy is tipped to an angle \(\theta\) from the vertical. [You need to find the height of the CM as a function of \(\theta\). It helps to think first about the height of the hemisphere's center 0 as the toy tilts.]
(b) For what values of R and h is the equilibrium at \(\theta=0\) stable?
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QUESTION:
Figure 4.25 shows a child's toy, which has the shape of a cylinder mounted on top of a hemisphere. The radius of the hemisphere is R and the CM of the whole toy is at a height h above the floor.
(a) Write down the gravitational potential energy when the toy is tipped to an angle \(\theta\) from the vertical. [You need to find the height of the CM as a function of \(\theta\). It helps to think first about the height of the hemisphere's center 0 as the toy tilts.]
(b) For what values of R and h is the equilibrium at \(\theta=0\) stable?
ANSWER:Step 1 of 2
(a)
The gravitational potential energy is given as,
\(U=m g H\)
Here, m is the mass, g is the acceleration due to gravity and H is the height above the center.
The gravitational potential energy is the same as for the point mass at the center of the toy.
When the toy is tipped to an angle \(\theta\) from the vertical, the height of the CM above origin is changes from \((h-R)\) to \((h-R) \cos \theta\)
The height of the center of mass after tipped the toy is given as,
\(H=(h-R) \cos \theta+R\)
Substitute \((h-R) \cos \theta+R\) for H into equation (1.1),
\(U=m g[(h-R) \cos \theta+R]\)
Therefore, the gravitational potential energy when the toy is tipped to an angle \(\theta\) from the vertical is \(m g[(h-R) \cos \theta+R]\)
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