A 1200-kg car is being raised over water by a winch. At the moment the car is 5.

Question

A 1200-kg car is being raised over water by a winch. At the moment
the car is 5.0 m above the water (Figure 9-56), the gearbox breaks—allowing the
winch drum to spin freely as the car falls. During the car’s fall, there is no
slipping between the (massless) rope, the pulley wheel, and the winch drum. The
moment of inertia of the winch drum is 320 kg·m2, and the moment of inertia of
the pulley wheel is 4.00 kg·m2. The radius of the winch drum is 0.800 m, and the
radius of the pulley is 0.300 m. Assume the car starts to fall from rest. Find the
speed of the car as it hits the water. solve thir problem using torque and forces, NOT energy! please help!

Explanation / Answer

Ok so first you need to calculate the Torques of each rotational mass. Recall that or the cross product of the radius and the rotational force of each object now also recall that rotational force is: F=I where I is the moment of inertia for the object and is its rotational acceleration

Intially the system is at rest so both Torques are 0Nm or J and the car only has gravitational potential energy or mgh so Eo=Two+Tpo+Uc where Uc=the potential energy of the care

Finally the system can be represented by Ef=Tw+Tp+(1/2)(m)(v2) so we want to solve for v which is the velocity at which the car hits the water. Now Eo=Ef so set both equations equal to each other.

Two+Tpo+Uc=Tw+Tp+(1/2)(mc)(v2) now Two and Tpo are both 0 so:

Uc=Tw+Tp+(1/2)(mc)(v2) :now isolate v:

[(2)(Uc-Tw-Tp)/(mc)]=v

Now we need to find the Torques we already have the radii for each rotational body so now we need the F which is given by F=I and I is given so we just need to find . We can use the identity =(a/r) or r=a.

Now we can use newtons 2nd law as it applies to rotational motion: Sum up the forces on each object

Fc=mca => mcg-T1=mca , Fp=I=(1/2)(m)(r2)(a/r) or (1/2)(mp)(r)(a) => T1-T2=(1/2)(mp)(rp)(a), Fw=(1/2)(mw)(rw)(a) => T2=(1/2)(mw)(r)(a) so now add all of the equations together which yields:

mcg-T1+T1-T2+T2=(mc+(1/2)(mp)(rp)+(1/2)(mw)(rw))a Notice all the tensions cancel out.Now to solve for a we need to solve for the mass of the pulley and winch which we can do using the moment of inertia for each one: Iw=(1/2)(mw)(rw)2 or 2(Iw)/(rw)2=(mw) => mw=(2)(320)/(0.800)2=1000 and mp=2(Ip)/(rw)2=2(4)/(0.300)2=88.89 Now plug in the masses to solve for a:

(mcg)/(mc+(1/2)(mp)(rp)+(1/2)(mw)(rw))=a

a=((1200)(9.8))/(1200+(.5)(88.89)(0.300)+(.5)(1000)(0.800))=a=7.289m/s2 Now solve for the torque of each rotational mass. =(a/r) so Tw=Iw(a/rw)=320(7.289/0.800)=2916J and Tp=Ip(a/rp)=4(7.289/0.300)=95.85J Now use the relation [(2)(Uc-Tw-Tp)/(mc)]=v to solve for v: [(2)(mcgh-Tw-Tp)/(mc)]=v

[(2)((1200)(9.8)(5.0)-(2916)-(95.85))/(1200)]=v=9.643m/s using 3 significant figures 9.64m/s

You have to use energy in this question because Torque is not a force it is an energy quantity since it is a force times a distance and the unit is Ns but you also have to use for to solve for

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