As part of an experiment in physics lab, small metal ball of radius r = 2.1 cm r

Question

As part of an experiment in physics lab, small metal ball of radius r = 2.1 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 3.9 m. The ball is solid with a uniform density and a mass M = 347 g.

1)

How high above the top of the loop must it be released in order that the ball just makes it around the loop? (answer in m)

2)

Now instead of a sphere, what if we rolled a solid disk with the same mass and radius. How high above the top of the loop must it be released in order that the disk just makes it around the loop? (answer in m)

3)

Finally, what if we had allowed a block with the same mass as the sphere to slide on the ramp. The block slips without friction but does not rotate. How high above the top of the loop must it be released in order that the block just makes it around the loop? (answer in m)

Explanation / Answer

1) M = 347 g = 0.347 kg   R = 3.9 m.   r = 2.1 cm = 0.021 m

take height = h.

take velocity at the hightest point = v.

mv^2/(R-r)=mg

v = (g*(R-r))^0.5

so

moment of inertia of a sphere I = (2/5)mr^2

conservation of energy

mgh= (1/2)mv^2+(1/2)Iw^2+mg(2R)

gh= (1/2)v^2+(1/2) (2/5)r^2w^2+g(2R)

h =2R , v=wr

9.8*((2*3.9)+x)=0.5*9.8*(3.9-0.021)+0.5*0.4*9.8*(3.9-0.021)+9.8*2*3.9

x=2.7153 m

2)

moment of inertia of disk is (1/2)mr^2

the conservation of energy,

9.8*(7.8+x)=0.5*9.8*(3.9-0.021)+0.5*0.5*9.8*(3.9-0.021)+9.8*2*3.9

or x=2.91 m

3) the conserving energy,

mgh=0.5mv^2+mg*2R and mv^2/R=mg >> mv^2 = mgR

h=0.5mgR+2R   

(7.8+x)=0.5*3.9+2*3.9

x=1.95 m

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