After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands

Complete question is;

After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.

Tangential speed of foot just before it lands is; v = 5.37m/s

Explanation:

Let U (potential energy) be zero on the ground.

So, initially, U = mgh

where, h = 0.98/2 = 0.49m (midpoint of the leg)

Now just before the leg hits the floor it would have kinetic energy as;

K = ½Iω²

where ω = v/r and I = ⅓mr²

So, K = ½(⅓mr²)(v/r)²

K = (1/6) × (mr²)/(v²/r²)

K = (1/6) × mv²

From principle of conservation of energy, we have;

Potential energy = Kinetic energy

Thus;

mgh = (1/6) × mv²

m will cancel out to give;

gh = (1/6)v²

Making v the subject, we have;

v = √6gh

v = √(6 × 9.81 × 0.49)

v = √28.8414

v = 5.37m/s

no. dot product   =0 when vecs at right angles