Question

A 1500 kg car starts from rest and drives around a flat 50-m-diameter circular track. The forward force provided by the car’s drive wheels is a constant 1000 N.
a. What are the magnitude and direction of the car’s acceleration at t = 10 s? Give the direction as an angle from the r-axis.
b. If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?

Short answer

a) Acceleration is 1.9 m/sec² at 21^o from r axis.

b) Car will start sliding after 8.25 sec

Step-by-step solution

Part(a) Step 1: Given information

Mass of the car, m=1500 kg
Diameter of the circular track, d=50 m
The forward force, F=1000 N
Time, t=10 sec

Part(a) Step 2: Explanation

We can calculate tangential acceleration as

a_t =F/M ………..................(1)

Where F= forward force, m = mass of the car
And we can calculate radial acceleration as

a_rad =v²/r ………...............(2)

Where v=velocity, r = radius of the track

We can calculate acceleration

$a=\sqrt{{a_t}^2+{a_r}^2}$

and angle of acceleration is

$\theta ={\tan }^{-1}\left(\frac{a_t}{a_r}\right)$

Substitute the value to get tangential acceleration in (1)

$a_t=\frac{1000N}{1500kg}=0.66m/s^2$

Velocity of the car after 10 sec is

$v=0+(0.66m/s^2)\times 10s=6.6m/s$

Similarly get radial acceleration by substituting values in (2)

role="math" localid="1649086630580" $a_r=\frac{{(6.6m/s)}^2}{25m}=1.77m/s^2$

Now calculate acceleration

$a=\sqrt{(a_t^2+a_r^2)}=\sqrt{{(0.66m/s^2)}^2+{(1.77m/s^2)}^2}=1.9m/s^2$

Direction of acceleration is

(the angle is)

$\theta ={\tan }^{-1}\left(\frac{0.66m/s^2}{1.77m/s^2}\right)={21}^o$

Part(b) Step 1 : Given information

Mass of the car, m=1500 kg
Diameter of the circular track, d=50 m
The forward force, F=1000 N
Time, t=10 sec

Part(b) Step 2 : Explanation

First calculate the velocity of the car using

$$\begin{gathered} \frac{m{v}^2}{r}=mg \\ v=\sqrt{rg} \end{gathered}$$

Substitute the values we get

$v=\sqrt{\left(25m\right)(9.8m/s^2)}=15.67m/s$

The car will start to slide when the final velocity v will become 0

$$\begin{gathered} v=u+at \\ Substitutevalues \\ 0=15.67m/sec-(1.9m/s^2)t \\ solvefort \\ t=8.25sec \end{gathered}$$