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Phantasy [73]
3 years ago
7

NIL

Physics
1 answer:
rosijanka [135]3 years ago
3 0

(a) Centripetal: 8.47 m/s^2, tangential: 1.35 m/s^2

The radius of the track (which is the distance of the car from the centre of the circular path) is

r = 0.30 km = 300 m

The angular acceleration is

\alpha = 4.5\cdot 10^{-3}rad/s^2

We can find the angular velocity of the car after half of a lap using the equivalent of the SUVAT equation for rotational motions:

\omega_f^2 - \omega_i^2 = 2\alpha \theta

where

\omega_i = 0 since the car starts from rest

\theta = \pi is the angular displacement after half a lap

Solving for \omega_f,

\omega_f = \sqrt{2\alpha \theta}=\sqrt{2(4.5\cdot 10^{-3})(\pi)}=0.168 rad/s

Now we can find the centripetal acceleration with the formula:

a_c = \omega^2 r = (0.168)^2 (300)=8.47 m/s^2

while the tangential acceleration is simply given by

a_t = \alpha r = (4.5\cdot 10^{-3})(300)=1.35 m/s^2

(b) The mass of the car is not given, so it is not possible to find the forces.

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