Question

Consider a runner in lane 1 with a speed v1 and a runner in lane 4 with a speed v4 , where the ratio of the speeds v4/v1 is 2 . The radius of lane 4 is b times the radius of lane 1, and both runners have the same mass. Which of the following claims is correct about the net forces F1 and F4 exerted on the runners in lanes 1 and 4, respectively?

a. F4 is equal to 2/b F1.
b. F4 is equal to 4/b F1.
c. F4 is equal to 2/b^2 F1.
d. F4 is equal to 4/b^2 A F1.

Answer 1

Answer:

$F_4=\frac{4}{b}F_1$

Explanation:

Centripetal force is the net force acting on a body which makes it move along a curved path. This force is always towards the center of curvature.

The centripetal force is given by:

F = mv² / r

where m is the mass of the body, v is the velocity of the body, r is the radius, F is the centripetal force and v²/r is the centripetal acceleration.

Given that:

$\frac{v_4}{v_1}=2\\\\v_4=2v_1\\\\ Also, radius\ of\ lane\ 4(r_4)=2*radius\ of\ lane\ 1\\\\Let\ r_1\ be\ radius\ of\ lane\ 1,m=mass\ of\ runner\ in\ lane\ 1=mass\ of\ runner\ in\ lane\ 4\ \\and\ v_4=velocity\ of\ runner\ in\ lane\ 4.\ Hence:\\\\r_4=b*r_1=br_1\\\\The\ centripetal\ force\ for\ lane\ 4(F_4)\ is:\\\\F_4=\frac{mv_4^2}{r_4}\\\\ F_4=\frac{m(2v_1)^2}{br_1}\\\\F_4= \frac{4mv_1^2}{br_1} \\\\But\ F_1=\frac{mv_1^2}{r_1}\\\\Hence\ F_4=\frac{4}{b}F_1$

Answer 2

F4 is equal to 4/b F1. A further explanation is provided below.

According to the question,

• $\frac{v_4}{v_1} =2$
• $\frac{r_4}{r_1} = b$

The net force of the object in uniform circular motion will be:

→ $F = \frac{mv^2}{r}$

∴ $F_1 = \frac{m_1 v_1^2}{r_1}$

  $F_4 = \frac{m_4 v_4^2}{r_4}$

Now,

→ $\frac{F_1}{F_4} = \frac{m_1}{m_4}\times (\frac{v_1}{v_4} )^2\times \frac{r_4}{r_1}$

   $\frac{F_1}{F_4} = \frac{1}{4}\times b$

   $F_1 = \frac{b}{4} F_4$

   $F_4 = \frac{4}{b} F_1$

Thus the above response i.e., "option b" is correct.

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