A non-uniform round object of mass M and radius R is found to have its total kinetic energy to be given by KE =
Mv² when rolling down an incline.What is this object's moment of inertia?

Question

A non-uniform round object of mass M and radius R is found to have its total kinetic energy to be given by KE =

Mv² when rolling down an incline.What is this object's moment of inertia?

Quizplus · Accepted Answer

Using the formula for kinetic energy of a rolling object, KE = (1/2) Iω^2 + (1/2)mv^2 where I is moment of inertia, ω is angular velocity, and v is linear velocity. Assuming the object is rolling without slipping, we have v = Rω Substituting this into the above equation, we get KE = (1/2) Iω^2 + (1/2)mR^2ω^2 Simplifying, KE = (1/2) (I + mR^2)ω^2 Comparing this with the given formula for KE, we can write I + mR^2 = ea800b_1901_90f6_846e_05e9d2843d58_TB6864_11 MR² We are given the mass and radius of the object, but we need to find the moment of inertia. Let's use the parallel axis theorem, which states that the moment of inertia of an object about any axis parallel to its center of mass is given by I = Icm + md^2 where Icm is moment of inertia about the center of mass, m is mass, and d is distance of the parallel axis from the center of mass. Let's choose the axis of rotation to be at the base of the incline where the object starts rolling. At this point, the linear velocity is zero, so all the kinetic energy is rotational. Then, we have KE = (1/2) Iω^2 Substituting the given formula for KE and the formula for I using the parallel axis theorem, we get (1/2) (Icm + md^2)ω^2 = 11ea800b_1901_90f6_846e_05e9d2843d58_TB6864_11 MR² Simplifying, Icmω^2 + md^2ω^2 = 22ea0017_1901_e374_e9f1_8c01956a1c73_TB6864_11 MR² Since the object is rolling without slipping, we have v = Rω Differentiating both sides with respect to time, we get a = Rα where a is linear acceleration and α is angular acceleration. But we also have a = gsinθ where θ is the angle of incline. Therefore, Rα = gsinθ Substituting α = a/R and rearranging, we get ω^2 = (2/3) (g/R) sinθ Substituting this into the previous equation, we get Icm (2/3) (g/R) sinθ + md^2 (2/3) (g/R) sinθ = 22ea0017_1901_e374_e9f1_8c01956a1c73_TB6864_11 MR² Simplifying and solving for Icm, we get Icm = (3/2) (22ea0017_1901_e374_e9f1_8c01956a1c73_TB6864_11 MR² - md^2 (2/3) (g/R) sinθ) Substituting the given values of mass and radius, as well as the acceleration due to gravity and the angle of incline, we get Icm = 11ea800b_1901_90f8_846e_2dcaeeaec10b_TB6864_11 MR² Finally, using the parallel axis theorem again, we can find the moment of inertia about the base of the incline: I = Icm + md^2 = 11ea800b_1901_90f8_846e_2dcaeeaec10b_TB6864_11 MR² + M(R/2)^2 = 11ea800b_1901_90f8_846e_2dcaeeaec10b_TB6864_11 MR² Therefore, the answer is B.

A Non-Uniform Round Object of Mass M and Radius R