A sphere and a disk, each having the same mass and radius, are released together

Question

A sphere and a disk, each having the same mass and radius, are released together at the top of an inclined plane and roll without slipping and with negligible rolling friction (in other words, an ideal conservative system). (a) Prove/Derive that v_sphere = 1.07 v_disk (b) If they descend a height of 3.0 meters over a horizontal distance of 4.0 meters, calculate how much longer the cylinder takes to reach the bottom than the sphere. The beam below is perfectly level (horizontal), has length 4m, and mass 10kg. The box which weighs 40N is 1m from the left side. Find the angle of the rope on the left.

Explanation / Answer

2) Applying energy conservation,

PEi + KEi = PEf + KEf

m g h + 0 = m v^2 /2 + (2 m r^2 / 5) (v_sphere / r)^2 /2

m g h =m v_sphere^2 /2 + m v_sphere^2 / 5 = 7m v_sphere^2 / 10

and

m g h = (m r^2 /2 ) (v_disk / r)^2 / 2

m gh = m v^2 /2 + m v_disk^2 / 4

m g h = 3 m v_disk^2 / 4

So,

3 m v_disk^2 / 4 = 7 m v_sphere^2 / 10

v_sphere^2 = 1.07 v_disk^2

v_sphere = 1.035 v_disk

So i disapprove this.

(b) For sphere,

m g h = 7 m vs^2 / 10

vs = sqrt(10 g h / 7) = 6.48 m/s

For cylinder,

m g h = 3 m vc^2 / 4

vc = sqrt(4 x 9.8 x 3 / m) = 6.26 m/s

L = sqrt(3^2 + 4^2) = 5 m

vf^2 - vi^2 = 2 a d

6.48^2 = 2 as (5)

as = 4.199 m/s^2

ts = 6.48 / 4.199 = 1.543 s
ac = 6.26^2 / 2 x 5 = 3.919 m/s^2

tc = 1.597 s

t = tc - ts = 0.054 sec

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