Hope this helps.

=> v^2 = r*g*tan(theta)

does that make sense the smallest speed they can move without relying on friction friction isnt a concern at low speeds, right or do you mean the car would fall off the bank i think you need friction no matter what.

Fg(down slope) = mg*sin(theta)

You can do the same at the minimum radius, and solve for the minimum speed:

The component of gravity pulling the car down the slope is:

Fc(up slope) = mv^2/r*cos(theta)

Fg(down slope) = mg*sin(theta)

Hope this helps.

You can do the same at the minimum radius, and solve for the minimum speed:

v= [r*g*tan(theta)]^0.5 = [(112m)*(9.8m/s^2)*(0.3396)]^0.5

=> v^2 = r*g*tan(theta)

v= [r*g*tan(theta)]^0.5 = [(112m)*(9.8m/s^2)*(0.3396)]^0.5

On A Banked Race Track

Fc(up slope) = mv^2/r*cos(theta)

mv^2/r*cos(theta) = mg*sin(theta)

On A Banked Race Track

jus make it more clear. (theta)= 165-112=53m. apply tan to finds the angle. Fc = F(theta). F(theta)= mgsin(theta). Fc= (mv^2/r). then mass cancel and you would be left with V^2= (rgsin(theta))/(cos(theta)). square root both sides and you will find the answer. for max use the bigger r and for mini use the smaller.

jus make it more clear. (theta)= 165-112=53m. apply tan to finds the angle. Fc = F(theta). F(theta)= mgsin(theta). Fc= (mv^2/r). then mass cancel and you would be left with V^2= (rgsin(theta))/(cos(theta)). square root both sides and you will find the answer. for max use the bigger r and for mini use the smaller.

does that make sense the smallest speed they can move without relying on friction friction isnt a concern at low speeds, right or do you mean the car would fall off the bank i think you need friction no matter what.

= 23.4m/s maximum speed.

The component of gravity pulling the car down the slope is:

This post is last updated on hrtanswers.com at Date : 1st of September – 2022