Question

A police officer in hot pursuit of a criminal drives her car through an unbanked circular (horizontal) turn of radius 300 m at a constant speed of 22.2 m/s. Her mass is 55.0 kg. To the nearest degree, what is the angle (relative to vertical) of the net force of the car seat on the officer?

Answer 1

Answer:

The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

Explanation:

Given:

Mass of the driver is, $m=55\ kg$

Radius of circular turn is, $R=300\ m$

Linear speed of the car is, $v=22.2\ m/s$

Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.

The forces are:

1. Weight = $mg=55\times 9.8=539\ N$

2. Centripetal force, 'F', which is given as:

$F=\frac{mv^2}{R}\\F=\frac{55\times (22.2)^2}{300}\\\\F=\frac{55\times 492.84}{300}\\\\F=\frac{27106.2}{300}=90.354\ N$

Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,

$\tan \theta=\frac{90.354}{539}\\\tan \theta=0.1676\\\theta=\tan^{-1}(0.1676)=9.52\approx 10$°

Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.