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