Question

You are on a circular racetrack that does not have banked turns. The radius of the tracks circle is 2.5 x 10^2m. Starting from rest you accelerate uniformly at a rate of 4m/s^2. The coefficient of friction is .8, how many seconds will it take to start sliding off of the track?

Answer 1

Answer on Question #48381, Physics, Mechanics | Kinematics | Dynamics

You are on a circular racetrack that does not have banked turns. The radius of the tracks circle is $2.5 \times 10^{2} \mathrm{~m}$. Starting from rest you accelerate uniformly at a rate of $4 \mathrm{~m} / \mathrm{s}^{2}$. The coefficient of friction is .8, how many seconds will it take to start sliding off of the track?

Solution:

Centripetal acceleration is part of moving in a circular path. Centripetal acceleration points toward the center of the circular path of the car, but is felt by passengers as a force pushing them to the outer edge of the circular path.

The equation for centripetal acceleration is:

$$a _ {r} = \frac {v ^ {2}}{r}.$$

The dynamics equation is

$$m a _ {r} = F _ {f r}$$

The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:

$$F _ {f r i c t} = \mu F _ {n o r m} = \mu m g$$

Thus,

$$m a _ {r} = \mu m g$$ $$a _ {r} = \mu g$$ $$\frac {v ^ {2}}{r} = \mu g$$

The maximum velocity is

$$v = \sqrt {\mu g r} = \sqrt {0 . 8 * 9 . 8 * 2 5 0 0} = 1 4 0 \mathrm {m / s}$$

The time we obtain from kinematics equation:

$$t = \frac {v - v _ {0}}{a} = \frac {1 4 0 - 0}{4} = 3 5 \mathrm {s}$$

Answer: $t = 35$ s

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