A solid, uniform disk of radius 0.250 m and mass 65.0 kg rolls down a ramp of le

Question

A solid, uniform disk of radius 0.250 m and mass 65.0 kg rolls down a ramp of length 3.90 m that makes an angle of 14.1° with the horizontal. The disk starts fromrest from the top of the ramp. (a) Find the speed of the disk's center of masswhen it reaches the bottom of the ramp.
1 m/s

(b) Find the angular speed of the disk at the bottom of theramp.
2 rad/s (a) Find the speed of the disk's center of masswhen it reaches the bottom of the ramp.
1 m/s

(b) Find the angular speed of the disk at the bottom of theramp.
2 rad/s

Explanation / Answer

intial energy of the disk = Mgh M = mass of the disk h = height = L tan . L = length of the slope as the disk reached the bottom of incline to pocessrotational and transilatory kinetic enregys                    Mgh = Mv^2 / 2 + I^2 / 2 .           = V / r                        Mgh= MV^2 / 2 + I(V/r)^2 / 2 .    from the above equation we can solve for linearvelocity v of the disk    from the above equation we can solve for linearvelocity v of the disk                        V = linear velocity of the disk .                        r = radius of the disk .                         I= mr^2 / 4                         m = mass of the disk                                                       

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