Learnohub

Question:

a motorcyclist loops a vertical loop of diameter 50 m without dropping down even at the highest point. what is the min speed at the lowest and the highest point.

Answer:

To find the minimum speed you require to complete the loop, let us consider the figure below-

To get the minimum required speed to make the loop, at the top of the loop we require the resultant force to be 0. That is N=mg, where N is the centripetal force, which is given by N=mv²r where r is the radius of the circle. Thus,

mv²r = mg

So, v_t²/r = g

v_t² = gr

v_t = √ gr

Thus, minimum velocity at the top of the loop, v t .

The total energy at the top of the loop is equal to the potential energy at the top plus the kinetic energy at the top, respectively these are-

P E t = m g h = 2 m g r,

K E t = 1/ 2 m v t 2 = 1/ 2 m g r .

Thus the total energy at the top is:

E t = 2 m g r + 1/ 2 m g r, = 5/ 2 m g r .

In the above question, r = 50/2 = 25 m,

So E t = 5/2 x m x 9.8 x 25 = 612.5 m

We now find the total energy at the bottom, since there is no potential energy, this is simply the kinetic energy,

E b = 1/ 2 m v b 2 .

Where vb is the speed we are looking for. With our assumptions, the energy at the top (Et) equals the energy at the bottom (E_b) giving

E b = Et

So, 1/2 mv b 2 = 5/2 mgr

mv b 2 = 5 mgr

v b 2 = 5 gr

v b = \sqrt5 gr

Thus, v_b = √5 x 9.8 x 25

= 35 m/s.

Not what you are looking for? Go ahead and submit the question, we will get back to you.