# A Solid sphere of mass m and radius r is gently placed in conveyer belt moving with speed v . if coefficient of friction is 2/7 distance travelled by sphere before pure rolling starts ?now when the sphere is put on the belt does it instantaneously move with the same linear speed on the belt or does it have an angular velocity too ?pls explain elaborately !!thanks

33 Points
10 years ago

linear eq. for sphere-

Fr = ma

=> (2mg)/7 = ma

=> (2g)/7 = a

Rotational eq.-

Fr.r = Iα

=> (2mg.r)/7 = (2mr2.α)/5

=> g/7 = (r.α)/5

=> α = 5g/7r

For pure rolling we know that velocity of bottom most point = velocity of the platform (i.e. the conveyer belt, Vp)

so, Vcm + Vζ = Vp

=> (u + at) + (ω0 + αt)r = Vp

=> (0 + 2gt/7) + (0 + 5gt/7r)r = v

=> 2gt/7 + 5gt/7 = v

=> t = 7v/7g

=> t = v/10

when  the sphere will be put on to the belt, it wont start moving with the same linear speed, it will take some time.

frition will prodie this force that will make the ball to move linearly as well as rotate.

but at the instant when it will be placed it wont have any linear or angular velocity, it will slip

:)