Rolling ball down slope at specific time

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SUMMARY

A uniform solid ball with a diameter of 0.5m and a mass of 0.1kg rolls down a slope starting from rest. The relevant equations include the moment of inertia, given by I = 2MR²/5, and the relationship between linear velocity and angular velocity, v = Rω. The rotational kinetic energy is expressed as 1/2 * I * ω², which simplifies to Mv²/5. The height of the ball as a function of time is h = 1/2 * g * sin(θ) * t², allowing for the calculation of velocity as a function of time.

PREREQUISITES
  • Understanding of rotational dynamics and moment of inertia
  • Familiarity with energy conservation principles in physics
  • Basic knowledge of kinematics and equations of motion
  • Ability to manipulate algebraic equations involving trigonometric functions
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  • Study the derivation of the equations of motion for rolling objects
  • Learn about energy conservation in rotational systems
  • Explore the relationship between linear and angular velocity in depth
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Homework Statement


A uniform solid ball rolls down a slope. If the ball has a diameter of .5m and a mass of .1 kg find the following:

The equation which describes the velocity of the ball at any time, given that it starts from rest.


Homework Equations



I've tried using energy equations, thus: 1/2 Iw^2+1/2Mv^2+mgh = constant (Kf +Uf = Ki + Ui) but that doesn't have "t" as a variable. I have also tried θ-θi=1/2(ωi+ω)t. But that didn't end up taking the radius or mass into consideration.

The Attempt at a Solution



More or less stated in the relevant equations. This is due in a few hours, so anything is much appreciated. Thank you in advance!
 
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Moment of inertia of the sphere is 2MR^2/5 and v = R*w.
So the rotational kinetic energy is given by 1/2*I*w^2 = 1/2*2M*R^2/5*v^2/R^2 = M*v^2/5.
And h = 1/2*g*sin(theta)*t^2.
Substitute these values in the relevant equation to solve for t.
 

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