Question

A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass of m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.
(a) The are cannot make a turn without skidding.
(b) If the car turns at a speed less than 40 km/hr, it will slip down.
(c) If the car turns at the car is equal to mv2r.
(d) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than mv2r.

Answer 1

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Secondary School Physics 20 points

A circular road of radius r is banked for a speed v=40 km/h. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.

(a) the car cannot make a turn without skidding.

(b) if a car turns at a speed less than 40 km/h it will slip down.

(c) if the car turns at the correct speed of 40 km/h , the force by the road on the car is equal to mv²/r.

(d) if the car turns at the correct speed of 40 km/h , the force by the road on the car is greater than mg as well as greater than mv²/r.

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Answers

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tiwaavi

tiwaavi Genius

Answer ⇒ Option (b). and Option (d).

Explanation ⇒ If the Road will be banked for the speed of 40 km/hr, then for this speed, the angle of banking will be given by,

tanθ = v²/rg

Now, if the speed of the car will be same as the given speed then the car may turn without skidding. Hence, Option (a). can not always be correct.

If the car will turn at the speed less than the 40 km/hr then it will move inwards due to which it will slip down. Hence, (b). is correct.

If the car will turn at the exact speed of 40 km/hr then the Force exerted by the Road on Car will be mv²/rSinθ. Thus, Option (c), is incorrect.

If the car will turn at the speed of 40 km/hr then the Force exerted will be greater than mv²/r because the value of sine always less than 1 between 0° to 90°. Also, It will be greater than mg because, NCosθ = mv²/r

⇒ N = m²/rCosθ.

Also the Cosine is having values less than 1 between 0 ° to 90°, therefore, It will always be greater than that.

Hence, Option (d) is correct.

Hope it helps.

Answer 2

A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass of m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.  If the car turns at a speed less than 40 km/hr, it will slip down and If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than mv2r.

Explanation:

• In the event that the Road will be banked for the speed of 40 km/hr, at that point for this speed, the edge of banking will be given by,

$\tan \theta=\frac{v^{2}}{r g}$

Presently, if the speed of the vehicle will be same as the given speed then the vehicle may turn without slipping. Thus, Option (a). can't generally be right.

• In the event that the vehicle will turn at the speed not exactly the 40 km/hr then it will move inwards because of which it will descend. Consequently, Option (b) is right.

• In the event that the vehicle will turn at the accurate speed of 40 km/hr then the Force applied by the Road on Car will bemv²Sinθ/r. Along these lines, Option (c), is erroneous.  

• On the off chance that the vehicle will turn at the speed of 40 km/hr then the Force applied will be more noteworthy than (mv^2)/r   in light of the fact that the estimation of sine in every case under 1 between 0° to 90°. Likewise, It will be more prominent than mg on the grounds that,    

$N \cos \theta=\frac{m v^{2}}{r}$

$N=\frac{m^{2} \cos \theta}{r}$

Likewise the Cosine is having values under 1 between 0 ° to 90°, in this way, It will consistently be more noteworthy than that. Consequently, Option (d) is right.

Thus, Option (b) and (d) are correct.