A child of mass 5 kg is going round a merry-goround that makes 1 rotation in 3.14 s. The radius of the merry-go-round is 2 m. The centrifugal force on the child will be

Question

A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

Answer 1

To calculate the centrifugal force acting on the child going around the merry-go-round, we need to use the formula for centrifugal (or centripetal) force:

F = m \cdot a_c

where F is the centrifugal force, m is the mass of the child, and a_c is the centripetal acceleration.

The centripetal acceleration a_c can be calculated using the formula:

a_c = \omega^2 \cdot r

where \omega is the angular velocity in radians per second, and r is the radius of the merry-go-round.

First, we need to find the angular velocity \omega. Given that the time for one complete rotation is T = 3.14 seconds, we have:

\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14}

Approximating \pi as 3.14, we find:

\omega = \frac{6.28}{3.14} \approx 2 rad/s

Now, we calculate the centripetal acceleration:

a_c = \omega^2 \cdot r = 2^2 \cdot 2 = 4 \cdot 2 = 8 \text{ m/s}^2

Then, using the mass of the child m = 5 kg, we find the centrifugal force:

F = m \cdot a_c = 5 \cdot 8 = 40 \text{ N}

Thus, the centrifugal force acting on the child is 40 N.

Therefore, the correct answer is:

40 N

Answer 2