A solid, uniform disk of radius 0.250 m and mass 63.4 kg rolls down a ramp of le

Question

A solid, uniform disk of radius 0.250 m and mass 63.4 kg rolls down a ramp of length 3.70 m that makes an angle of 12.9° with the horizontal. The disk startsfrom rest 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

         Given thatthe radius of disk is R = 0.250 m          mass of diskis M = 63.4 kg        Length fo theplane is L = 3.70m          The angle ofinclination is = 12.90          Initialvlocity is U = 0 m/s    ---------------------------------------------------------------------------------         From the given datathe height of plane is H = Lsin = ---------- m         Apply conservationof energy at top and bottom of incline then                   mgH = (1/2)I2 +(1/2)mV2           (since initial velocity is zero)                   mgH = (1/2)I(V/R)2+ (1/2)mV2        But the moment of inertiaof silid disk is I = MR2/2                   mgH = (1/2)(mR2/2)(V/R)2+(1/2)mV2                   mgH = (1/2)(m/2)(V)2+(1/2)mV2                      mgH= (1/2)[(m/2)+ m]V2                             gH = [(3/4)]V2                                     V = (4gH / 3 )1/2                             =--------- m/s This is the linear velocity at bottom The angular velocity is = V/R                                        =--------- rad/s                                      =--------- m/s This is the linear velocity at bottom The angular velocity is = V/R                                        =--------- rad/s         

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