Question

A student builds and calibrates an accelerometer and uses it to determine the speed of her car around a certain unbanked highway curve. The accelerometer is a plumb bob with a protractor that she attaches to the roof of her car. A friend riding in the car with the student observes that the plumb bob hangs at an angle of 15.0° from the vertical when the car has a speed of 21.5 m/s. (a) What is the centripetal acceleration of the car rounding the curve? m/s2 (b) What is the radius of the curve? m (c) What is the speed of the car if the plumb bob deflection is 7.00° while rounding the same curve? m/s

Answer 1

Answer:

a) 9.46M m/s^2

b) 48.83m

c) 68.22m/s

Explanation:

The centripetal force must be equal to the horizontal component (x component) of the gravitational force over the plum bob. Thus, we have

$F_C=F_{xg}=Mgcos\theta=M\frac{v^2}{r}$  ( 1 )

where M is the mass of the car, v its speed, r the radius of the curvature and g the gravity constant.

a) By replacing we obtain:

$F_c=M(9.8m/s^2)(cos15\°)=9.46M\ m/s^2$   ( 2 )

where it is only necessary to put the mass of the car M in (2).

b) By canceling M in (2) and taking apart r we get:

$gcos\theta =\frac{v^2}{r}\\\\r=\frac{v^2}{gcos\theta}=\frac{(21.5m/s)^2}{(9.8m/s^2)cos15\°}=48.83m$

c) If the angle is 7° the speed is given by:

$v=\sqrt{gcos\theta r}=\sqrt{(9.8m/s^2)(cos7\°)(48.83m)}=68.22\frac{m}{s}$

hope this helps!!