Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A thick walled hollow sphere has outer radius R. It rolls down an inclined plane without slipping and its speed at the bottom is u. If the inclined plane is frictionless and the sphere s!ides down without rolling, its speed at the bottom will be 5u/ 4 . What is the radius of gyration of the sphere?
Initially, when the sphere is rolling down the include without any slipping, then let the angular velocity of the sphere at the bottom of the incline be w. No slipping condition, V = wRFurthermore, due to conservation of energyPE lost = KE gainedmgh = 1/2mv^2 + 1/2 *I*w^2 (Where h is the height of the incline and I is the moment of inertia of sphere around its center)Also, I = m*Rg ^2 where Rg is the radius of gyration and also v = wR (No slipping condition)Hence, mgh = 1/2mv^2 + 1/2 *m*Rg ^2 *v^2 /R^22gh = [1+(Rg/R)^2]*v^2 (1) Furthermore, under no rolling casemgh = 1/2m*(5v/4)^2v^2 = 32/25 ghPutting it in the equation 12gh = [1+(Rg/R)^2]*32/25 ghHence, Rg = 3/4RSo, the radius of gyration of the sphere around its center is 3/4R
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -