about rotational motion?

Question

9-65
A 1200-kg car is being unloaded by a winch. At the moment shown in figure 9-46, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 320 kg m^2 and that of the pulley is 4 kg m^2. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water.

9-92
A uniform solid sphere of radius r starts from rest at a height h and rolls without slipping along the loop-the-loop track of radius R as shown in Figure 9-56. (a) What is the smallest value of h for which the sphere will not leave the track at the top of the loop? (b) What would h have to be if, instead of rolling, the ball slides without friction?

Answer 1

The figure is required to calculate the actual solution. Since there is no figure, I will try to guess at the geometry of the problem and provide technique:

The way to solve this is conservation of energy. The car will lose potential energy m*g*h, which will get converted to translational kinetic energy of the car .5*m*v^2, and rotational kinetic energy of the pulley .5*Ip*wp^2 and rotational kinetic energy of the winch drum, .5*Id*wd^2.

h is the distance to the surface of the water. I assume that is shown in the figure.
g is gravity, 9.81 m/s^2
m is the mass of the car 1200 kg
v is the speed of the car at impact
Ip is the moment of inertia of the pulley 4 kg m^2
wp is the rotational speed of the pulley. I'll derive that in a moment
Id is the moment of inertia of the winch drum 320 kg m^2
wd is the rotational speed of the drum. I'll derive that in a moment.

Since there is no slipping of the cable, the speed of the car, v, is the translational speed of the cable over it's entire length.
wp=v/rp, where rp is the radius of the pulley, 0.30 m
wd=v/rd, where rd is the radius of the drum 0.80 m

plug it all in to get
m*g*h=.5*m*v^2+
.5*Ip*v^2/rp^2+
.5*Id*v^2/rd^2

solve for v

9-92

IN this case the speed of he sphere at the top must be great enough that the centripetal force will keep it in the loop.

using that, solve for v, that is minimum.

With v, again, using conservation of energy
m*g*(h-R)=.5*m*v^2+.5*I*r^2

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