If a certain car, going with speed v1, rounds a level curve with a radius R1, it is just on the verge of skidding. If its speed is now doubl

Question

If a certain car, going with speed v1, rounds a level curve with a radius R1, it is just on the verge of skidding. If its speed is now doubled, the radius of the tightest curve on the same road that it can round without skidding is:A) 2R1B) 4R1C) R1/2D) R1/4E) R1

in progress 0
Sapo 5 years 2021-08-01T12:30:08+00:00 1 Answers 412 views 1

Answers ( )

    1
    2021-08-01T12:31:46+00:00

    Answer:

    R_{2}=4R_{1}

    Explanation:

    Let’s use the centripetal force definition.

    So the first force will be:

    F_{1}=ma_{c}=m\frac{v_{1}^{2}}{R_{1}} (1)

    Here:

    • V1 is the tangential velocity  
    • R1 is the first radius
    • m is the mass of the car.

    Now, the final speed is double of the first one, so V2 = 2V1, Therefore the force will be:

    F_{2}=m\frac{v_{2}^{2}}{R_{2}}=m\frac{4v_{1}^{2}}{R_{2}} (2)

    If we match both equation (1) and (2), we will find R2.

    m\frac{v_{1}^{2}}{R_{1}}=m\frac{4v_{1}^{2}}{R_{2}}

    \frac{1}{R_{1}}=\frac{4}{R_{2}}

    Finally,

    R_{2}=4R_{1}  

    I hope it helps you!

         

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )