Question

a solid sphere of mass M and radius r a hollow cylinder of mass M and radius 2r a solid sphere of mass M and radius 3r and a hollow sphere of mass M and radius 4r are released at the top of an incline assuming that friction is just sufficient for rolling of hollow sphere which of these will reach bottom of the incline simultaneously​

Answer 1

Answer:

Using formula plz to solve

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Answer 2

Given :

A solid sphere of mass M and radius r

a hollow cylinder of mass M and radius 2r

a solid sphere of mass M and radius 3r

hollow sphere of mass M and radius 4r

To Find :

Which among them will reach bottom first when released simultaneously from top following rolling motion

Solution:

•Acceleration in rolling motion is

a = (gsinA)/[1 + (I/mR²) ]

•where g is acceleration due to gravity

A is angle of inclination

I is moment of inertia

R is radius of object

•For solid sphere of mass M and radius r

I = 2mR²/5 = 2Mr²/5

a = 5gsinA/7

•For hollow cylinder of mass M and radius 2r

I = mR²/2 = M(2r)²/2 = 2Mr²

a = 2gsinA/3

•For solid sphere of mass M and radius 3r

I = 2mR²/5 = 2M(3r)²/5 = 18Mr²/5

a = gsinA/3

•For hollow sphere of mass M and radius 4r

I = 2mR²/3 = 2M(4r)²/3 = 32Mr²/3

a = 3gsinA/5

•Since acceleration of solid sphere of mass M and radius r is greater among all

•Hence , solid sphere of mass M and radius r will reach bottom first .