A uniform solid sphere rolls along a horizontal frictionless surface at 35 m/s and makes a smooth transition onto a frictionless incline having an angle of 300. How far up the incline does the sphere roll when it comes to a momentary stop? Note for a sphere I= 2/5Mr2, where M is the mass of the sphere and r is the radius of the sphere.
I just answered this below and you mean 30 degrees
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To find out how far up the incline the sphere rolls when it comes to a momentary stop, we need to apply the conservation of energy principle.
The initial energy of the system is solely kinetic energy and can be calculated as:
KE_initial = 1/2 * I * ω_initial^2 + 1/2 * M * v_initial^2
Where:
KE_initial is the initial kinetic energy
I is the moment of inertia of the sphere (given as 2/5 * M * r^2)
ω_initial is the initial angular velocity of the sphere (can be calculated using v_initial = ω_initial * r)
M is the mass of the sphere
v_initial is the initial linear velocity of the sphere
Given in the problem, the initial linear velocity of the sphere, v_initial, is 35 m/s.
Now, the sphere comes to a stop at some height on the incline, so all its energy is potential energy at that point. The potential energy is given by:
PE_final = M * g * h
Where:
PE_final is the final potential energy of the sphere at the point of momentary stop
M is the mass of the sphere
g is the acceleration due to gravity (approximated as 9.8 m/s^2)
h is the height up the incline we are trying to find
According to the conservation of energy principle, the initial kinetic energy (KE_initial) should equal the final potential energy (PE_final). So we can equate them:
1/2 * I * ω_initial^2 + 1/2 * M * v_initial^2 = M * g * h
Now, we can substitute the expression for ω_initial in terms of v_initial and r:
1/2 * (2/5 * M * r^2) * (v_initial / r)^2 + 1/2 * M * v_initial^2 = M * g * h
Simplifying this equation will allow us to solve for h, the height up the incline:
(2/5 * M * r^2) * (v_initial / r)^2 + M * v_initial^2 = 2 * M * g * h
We can now substitute the given values for M, r, v_initial, g, and solve for h.