Question

A 5ft tall 50kg gymnast is walking along a 7m long solid wooden beam supported but two legs on either side set 1m from the ends. If she is carrying a hollow metal cylinder of uniform density 2.7 g/cm3 and 1m length under her arm such that 20% of its length is in front of her center of mass how close to the edge can she walk before equilibrium is lost? The inner radius of the cylinder is 3cm and the outer radius is 7cm.

Answer 1

First we have to calculate the mass of the cylinder.

Inner radius = 3 cm

Outer radius = 7 cm

Length = 1 m = 100 cm

Volume of the cylinder is

Density = 2.7 g/cm^3

So,

Mass = density*volume = 2.7*V = 33912 g = 33.912 kg

Now we have the person whose mass is 50 kg

When their combined center of mass crosses 1 m from the edge, the planck will fall down.

Now, since 20% of the mass is in the front, the center of mass of the rod will be

1*(0.5-0.2) = 0.3 m behind him.

Using

M1r1 = M2r2

And

r1+ r2 = 0.3 m

So,

r1 = M2(0.3-r1) /M1

r1 = 33.912*(0.3-r1) /50

r1 =0.203-0.67824r1

r1 = 0.203/(1+0.67824) = 0.121 m from the leg

This means a distance of 1-0.121 = 0.879 m from the edge