3. A car of mass m drives through a curve on a horizontally...

3. A car of mass m drives through a curve on a horizontally flat road at a constant speed v. The curve has a radius of curvature R. a) Draw a free body diagram for the car (from a side view). b) What is the magnitude of the frictional force that keeps the car moving in a circle of radius R with a uniform speed v? Write the answer in terms of the quantities given in the problem (m, R, and/or v) and, perhaps, constants (e.g., g, ?, ½ ...). c) If the car has a mass of 1500 kg, then what is the magnitude of the frictional force acting on the car that keeps it moving in a circle with a radius of 50.0 m with a uniform speed of 15.0 m/s? {6.75 kN} d) If the coefficient of static friction between the tires and the road is ?s, derive an expression for the maximum safe speed a car of mass m can drive through a curve of radius R with a uniform speed v without sliding. Write the result in terms of the given quantities (?s, m, and/or R) and constants. i) Does the maximum safe speed increase or decrease on an icy road compared to a dry road? ii) Does the maximum safe speed increase or decrease for a tighter curve (smaller radius of curvature)? iii) Does the maximum safe speed increase or decrease for a small car compared to a larger SUV?

Transcript

00:01 As shown in the diagram with the red color is a circular road, which is at the same level, but it is a circular road.

00:11 And this road and shown with the green color is the actual position where a car of mass m is moving with a velocity b.

00:24 It is a circular road means it is a part of a circle.

00:27 If we complete it like this, then this becomes a circle and sees the center of that circle.

00:35 And the radius of that circle is r like the actual circular path with the car is taking, which is shown with the green color.

00:45 This is the corresponding circle for that.

00:49 So the radius is r, position of car, right? so first we need to draw the free body diagram.

00:58 Now when car is moving in this direction, like when car comes here in this direction, the car does not want to continue in the circular path.

01:11 It wants to skid off like this.

01:15 When it turns, it means along with the circular path, it does not want to turn its velocity.

01:22 It just want to skid off.

01:25 Means out of the road.

01:27 So we need a force actually to keep the car in the circular path and that force is called centipital force.

01:35 It always acts towards the center of the circle.

01:38 Like in this car, the force is acting in this direction towards the center of the circle.

01:48 Now, centipital force formula is mv square upon rm is a mass of the car, v is the velocity of the car on the circular path and r is the radius of the curve.

01:59 Now, centipital force is not the basic force of the nature.

02:03 This force is always provided by some other force.

02:07 So in general, we can say whatever the force which adds towards the center of the circle is actually provides the centripetal force.

02:14 Now we can see if the car wants to skid off in this direction, in this direction, or we can say to skid, then force of static friction between the wheels of the car and the road means avoids the skid of the car so force of static friction acts in this direction it is in the opposite direction this this is the force of static friction which wants to prevent the motion of the car across the road like this the car wants to move in this direction and force of force of friction keeps the car the same track so it means the force of friction means provides the centipital force right so if you draw the free body diagram then this is the means ground level this is a part of the circular road car is here the weight of the car m g is acting downward and the non the normal reaction from the floor is acting upward but now one more force acting on the car because car won't to skid off in this direction.

03:28 So the force of friction acts in this direction here.

03:35 So this is a free body diagram.

03:39 And regarding the force of friction, we know that force of friction is equal to coefficient of static friction into normal reaction.

03:47 Because this is a force of static friction which does not want the wheels of the car to move across the road, say from towards this side.

04:01 As the car is not moving in the vertical direction, the normal reaction n is equal to mg.

04:08 So we can say that the force of static friction, we can call it f s, is equal to mu s into m g.

04:16 This is a free body diagram of answer for part a.

04:23 Now in part b, they are asking what is the right expression for magnitude of the frictional force.

04:31 Now we know that it is a frictional force which which provides the centripetal force.

04:39 So the frictional force is equal to centrifugal force and we know the formula for centripetal force is mv square upon r.

04:49 So means frictional force can be written in terms of mv are using this formula.

04:57 Now in part c if mass of the car is 1500 kg, and car is taking a turn on a road of radius 50 meter and the maximum speed at which the car can turn without skiing out of the road.

05:20 Like car will skid out of the road if the centripetal force becomes more than the frictional force.

05:27 If the antitimital force will surely depend upon the velocity...

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