Physics, 09.11.2019 03:31 naynay394

Auniform spherical shell of mass = 4.50 kg and radius = 8.50 cm can rotate about a vertical axis on frictionless bearings. a massless cord passes around the equator of the shell, over a pulley of rotational inertia = 3.00 × 10−3 kg ∙ m2 and radius = 5.00 cm, and is attached to a small object of mass m = 0.600 kg. there is no friction on the pulley’s axle; the cord does not slip on the pulley. what is the speed of the object when it has fallen 82.0 cm after being released from rest? use energy considerations.

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Answer from: Nelly6978

v = 1.41739 [m/s]

Explanation:

Let:

M = shell mass = 4.5 [Kg]

R = shell radius = 0.085 [m]

$I_{p}$ = pulley's rotational inertia = $3*10^{-3}$ [Kg*m^2]

r = radius of the pulley = 0.05 [m]

m = object's mass = 0.6 [Kg]

h = fall distance = 0.82 [m]

v = velocity of the block in [m]

as there is no losses due to friction we are able to start, so as soon as we let the system interact the block will fall a height h. such energy must be equal to the total kinetic energy of the system, let's draw our equation's energy:

potential energy of the block : mgh

kinetical energy of the block : $\frac{1}{2}*m*v^2$

kinetical energy of the pulley (only rotational) : $\frac{1}{2}*I_{p}*\omega^2$

now who is the angular velocity omega?, well the only thing that is making the pulley rotate is the block, and in the external part of the pulley, each point is moving with an angular velocity produced by v:

we have the relationships between these two, with the following equation:

$v = \omega * r =\omega = v/r$

kinetical energy of the shell (again only rotational ) : $\frac{1}{2}*I*\omega^2$

the moment of inertia of a spherical shell is $I = \frac{2}{3}*MR^2$

again, omega is obtained the same way, with the exception of the new radius which is R (instead of r):

$v = \omega * R =\omega = v/R$

Finally we replace and solve v:

we have so far:

$mgh = \frac{1}{2} mv^2 + \frac{1}{2} I_{p}*\frac{v^2}{r^2} + \frac{1*2}{2*3}*MR^2*\frac{v^2}{R^2}\\mgh = \frac{1}{2} mv^2 +\frac{1}{2} I_{p}*\frac{v^2}{r^2} +\frac{1}{3}Mv^2\\mgh =(\frac{1}{2} m+ \frac{1}{2*r^2} I_{p}+\frac{1}{3} M)*v^2$

finally we solve v and replace values:

$v^2 =\frac{mgh}{\frac{1}{2} m+ \frac{1}{2*r^2} I_{p} +\frac{1}{3} M}\\ v=\sqrt{\frac{mgh}{\frac{1}{2} m+ \frac{1}{2*r^2} I_{p} +\frac{M}{3}}}$

$v= \sqrt{\frac{0.6*9.8*0.82}{\frac{1}{2} 0.6+ \frac{1}{2*(0.05)^2}*3*10^{-3} +\frac{4.5}{3}}}$

finally v = 1.41739 [m/s]

Answer from: Quest

answer: what are you talking about?

explanation:

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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