A uniform spherical shell of mass M = 19.7 kg and radius R = 1.62 m rotates about a vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 1.02 kg m2 and radius r = 0.63 m, and is attached to a small object of mass m = 6.6 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 after it falls a distance h = 4.63 m from rest? Use energy considerations.
Assume that I1=2MR²/3 and ω1 are the momemt of inertia and
angular velocity of the sphere of radius R;
I and ω2 are the momemt of inertia and
angular velocity of the disc of radius r;
m and v are the mass and linear velocity of the load,
and ω1•R= ω2•r = v.
From the energy conservation law
m•g•h = m•v²/2 + I1•ω1²/2 + I2•ω2²/2 = m•v²/2 +{(1/2)•(2MR²/3) •(v/R)² } + 1/2•I•(v/r)² =
= m•v²/2 +Mv²/3 +I•v² / 2• r².
v =sqrt[m•g•h/(m/2 +M/3 + I/2r²)] = 5.18 m/s.
(check my calculations)
To find the speed of the object after it falls, we can use energy considerations. The energy of the system is conserved, so the potential energy lost by the object is equal to the increase in kinetic energy.
First, we need to calculate the potential energy lost by the object as it falls a distance h = 4.63 m. The potential energy lost can be calculated using the formula:
ΔPE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
Plugging in the values, we have:
ΔPE = 6.6 kg * 9.8 m/s^2 * 4.63 m
Next, we need to find the increase in kinetic energy of the object. Since the cord does not slip on the pulley, the object and the pulley have the same linear velocity. The kinetic energy increase of the object can be calculated using the formula:
ΔKE = 0.5 * m * v^2
where m is the mass of the object and v is the velocity.
Now, let's calculate the moment of inertia of the pulley using the formula:
I = 0.5 * m * r^2
where m is the mass of the pulley and r is the radius of the pulley.
Plugging in the values, we have:
I = 0.5 * 1.02 kg m^2 * (0.63 m)^2
Now, we can equate the potential energy lost to the increase in kinetic energy and solve for the velocity:
ΔPE = ΔKE
m * g * h = 0.5 * m * v^2 + 0.5 * I * (v / r)^2
Simplifying and rearranging the equation, we have:
v^2 = 2 * g * h / (1 + (I / (m * r^2)))
Plugging in the known values, we have:
v^2 = 2 * 9.8 m/s^2 * 4.63 m / (1 + (1.02 kg m^2 / (6.6 kg * (0.63 m)^2)))
Finally, we can solve for the velocity by taking the square root of both sides:
v = sqrt(2 * 9.8 m/s^2 * 4.63 m / (1 + (1.02 kg m^2 / (6.6 kg * (0.63 m)^2))))
Calculating this using a calculator, we find the speed of the object after falling a distance of 4.63 m is approximately 6.096 m/s.