7. Two identical disks with radius R, mass M, and rotationalinertia I = 1/2 MR^2

Question

7. Two identical disks with radius R, mass M, and rotationalinertia I = 1/2 MR^2 roll without sliding across ahorizontal floor with the same speed and then up a inclines. Disk A rollsup its incline without sliding. On the other hand, disk B rolls up a frictionless incline. Otherwise the inclines areidentical. Disk A reaches a height hA above the floor before rolling down again. a) (8 points) What is the height hB which disk B willreach? 7. Two identical disks with radius R, mass M, and rotationalinertia I = 1/2 MR^2 roll without sliding across ahorizontal floor with the same speed and then up a inclines. Disk A rollsup its incline without sliding. On the other hand, disk B rolls up a frictionless incline. Otherwise the inclines areidentical. Disk A reaches a height hA above the floor before rolling down again. a) (8 points) What is the height hB which disk B willreach?

Explanation / Answer

Using vCM= R * w for pure rolling motion,wehave K = (1/2)ICM * (vCM/R)^2 +(1/2)MvCM^2 or K = (1/2)[(ICM/R^2) + M] *vCM^2 the conservation of mechanical energy gives vCM = [2gh/1 + (ICM/MR^2)]^(1/2) the velocity of disk A above the floor before rolling downagain is v1 = [2kghA/1 +(ICM/MR^2)]^(1/2) or v1^2 = [2kghA/1 + (ICM/MR^2)]------------(1) similarly,the velocity of disk B above the floor beforerolling down again is v2 = [2ghB/1 + (ICM/MR^2)]^(1/2) or v2^2 = [2ghB/1 + (ICM/MR^2)]-------------(2) (v1/v2)^2 = {[2kghA/1 +(ICM/MR^2)]/[2ghB/1 + (ICM/MR^2)]} or (v1/v2)^2 = k * (hA/hB) or hB = k* hA * (v2/v1)^2 k is the coefficient of kinetic friction fordisk A when it is rolling up the incline.

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