What is the maximum speed at which a car can turn around a curve of 30 m radius on a level road?

Answer

Verified

Hint:For a car to make a successful turn on a curved road two forces acting on the car have to be equal. One is the centrifugal force and the other is the frictional force. If the centrifugal force is higher than the car will slide to the right and if the frictional force is higher the car will slide to the left and we will never attain the maximum speed.

Formula Used:

Frictional Force,${F_{Friction}} = \mu .{F_{Normal}}$ ,where $\mu $is the coefficient of friction, and ${F_{Normal}}$ is the normal force acting upon the two surfaces. Centrifugal force,${F_{Centrifugal}} = - \dfrac{{m{v^2}}}{R}$ ,where m=mass of the body, v= velocity, and R= radius of the circular motion.

Complete step by step answer:

The frictional force is the force that is generated by two surfaces that slide against each other when in contact. This force is mainly affected by the texture of the surface and the amount of force that are binding the two surfaces together.For example, if an object is placed over another object then the frictional force will be the weight of the first object. The total frictional force can be calculated using the formula,${F_{Friction}} = \mu .{F_{Normal}}$, where $\mu $is the coefficient of friction, and ${F_{Normal}}$ is the normal force acting upon the two surfaces.When a body is moving in a circular motion, then two forces come into play, the centripetal force and the centrifugal force. The centrifugal force is a force that acts on the body moving in a circular motion and is always directed outwards in a perpendicular direction. This force mainly arises due to the inertia of the body.The formula to calculate the centrifugal force is given by${F_{Centrifugal}} = \dfrac{{m{v^2}}}{R}$, where m=mass of the body, v= velocity, and R= radius of the circular motion.When the car is making the turn, two forces are working on it. The centrifugal force directed outwards and the frictional force directed inwards. For the car to make a successful turn both these forces have to be equal.Thus, we can write ${F_{Frictional}} = {F_{Centrifugal}}$.Replacing the values in the above equation we get,$\mu .{F_{Normal}} = \dfrac{{m{v^2}}}{R}$. Now, we know ${F_{Normal}} = mg$ , this will give us,$\mu .mg = \dfrac{{m{v^2}}}{R}$.Further equating we get,$\mu .g = \dfrac{{{v^2}}}{R}$. From, which we get,${v^2} = \mu gR$.Putting the values of the parameters and g= 9.8m/s, we get,$v = \sqrt {\mu gR} = \sqrt {0.4 \times 9.8 \times 30} = \sqrt {117.6} = 10.84$ .Thus, the maximum velocity that can be attained by the vehicle while making the turn is 10.84 m/s.

Thus, (B) is the right option.

Note:Here, we have to keep in mind the forces working in a circular motion. When making a turn the force working outward always has to be equal to the force working inward. Thus, equating both will surely give us our desired result.

A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the,

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

Given,

Mass of the helicopter, mh = 1000 kg

Mass of the crew and passengers, mp = 300 kg

Total mass of the system, m = 1300 kg

Acceleration of the helicopter, a = 15 m/s2 Using Newton’s second law of motion, the reaction force R,

R – mpg = ma

= mp(g + a) = 300 (10 + 15) = 300 × 25 = 7500 N The reaction force will be directed upwards, the helicopter is accelerating vertically upwards. According to Newton’s third law of motion, the force on the floor by the crew and passengers = 7500 N, directed downward.(b)

Using Newton’s second law of motion, the reaction force R’ experienced by the helicopter can be calculated as,

R' - mg = ma

= m(g + a) = 1300 (10 + 15) = 1300 × 25 = 32500 N The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.