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# Mechanics: Motion on Incline, Banking on Tracks, Projectile.

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Thank you so much. These are not jokes but are problems from a teacher wishing to have a little fun with his students.

https://brainmass.com/physics/circular-motion/mechanics-motion-incline-banking-tracks-projectile-149483

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Here is the solution attached.

a) Determine the angle of the incline?

The angle of the incline  can be determined by solving the right angled triangle with perpendicular equal to 100 ft and the base equal to 200 ft as

Or

b) What angle would the man have to lean forward to, in order to lose his center of gravity and fall down the slope?

I think it should be "in order not to lose his center of gravity and fall down the slope?"

As there is no friction, the forces acting on the man are his weight mg and the normal force of the incline N. the weight has no torque about his center of gravity but the normal reaction may have. To remain standing on the incline while slipping and not to topple the torque due to the normal force N must also be zero and for this he must remain normal to the incline and hence incline with the vertical by the same angle  as the slopes makes with horizontal. Hence to remain without falling on the slope he must lean forward by angle  = 26.5650 with the vertical. If the angle is less then this he will fall behind and if the angle is more will fall forward.

c) Assuming the man did fall and rolled into a ball, what speed or velocity would the man be rolling at the bottom of the slope?

As there is no force other then the component of his weight along the incline, and the man rolled in to a sphere, his center of mass will accelerate with g*sin and will acquire a kinetic energy equal to the potential energy lost and hence the speed of center of mass is given by applying the law of conservation of energy as (ignoring his height which is not given)

Gain in KE = loss in PE

Or ½ mv2 = mg h

Or m/s.

d) Is this angular velocity??

When the man rolls into a ball the normal reaction will pass through the center of the ball which is the center of mass and hence there is no torque and thus no angular velocity gain by the man. For a sphere to roll over an incline friction is essentially required otherwise it will slide without rolling.

Problem #2

15 ft.

30 ft.
> A 4000 lb. Race car travels at 200 miles per hour around a race track
> Assume a frictionless system

a) Determine the angle of the incline on the race track?

The angle of the incline is again determine as in the first problem as

Or

b) Determine the vertical and horizontal force (Fv & Fh) exerted by the race car on the track?

As there is no friction, the forces acting on the car are the normal reaction of the track N and the weight of the car mg.

The normal reaction of the track can be resolved in horizontal and vertical direction. The vertical component will be N*cos which will balance the weight of the car and hence

N*cos = mg
Hence the vertical force exerted by the car on the track will be

Fv = N*cos = mg = 4000 lb. ----------------------- (1)

The horizontal component of the normal reaction is given by (substituting from equation 1)

Fh = N*sin  = (Fv/cos) sin = Fv*cot = 4000*2 = 8000 lb.

Hence the horizontal force exerted by the car on the track will outward and equal in magnitude to 8000 lb.

c) Assuming no friction, is Fh sufficient to cause the race car to slide down the track?

This depends on the radius of the circular track.

The component of the normal force in horizontal direction is the only force acting on the car in that direction. This force is providing necessary centripetal force to move the car on the circular track. The centripetal force on a body of mass m moving with uniform speed v on a circular path of radius R is given by (mv2/R).

If the horizontal component is less then the required centripetal force (mv2/R), the radius of the path will increase and the car will slide up. But if the horizontal component is more then the required centripetal force (mv2/R), the radius of the path will decrease and the car will slide down.

arc??
Problem #3

V

60o

H

> A bullet weighs 10 grams and is fired from a gun at an angle of 60o and at an initial velocity of 500 meters/sec

This is the motion in two dimensions or motion in a plane.

The force acting on a projectile is the gravitational force and hence the acceleration is that of acceleration due to gravity which is vertically downwards.

As any vector is having no effect in its perpendicular direction, there will be no component of acceleration in horizontal direction and hence we can solve the problem for the horizontal and vertical motion separately and time will be linking parameter for the two.

Thus resolving initial velocity v0 = 500 m/s horizontally and vertically we have

Initial horizontal velocity vox = v0 cos 

Horizontal acceleration ax = 0

Initial vertically velocity voy = v0 sin 

Vertical acceleration ay = -g = - 9.8 m/s2

a) Calculate the height (V) the bullet will reach at theoretical zero velocity before it begins its descent?

As the final vertical velocity at the highest point of the path is zero, the maximum height attained V can be determined using the third equation of motion as

[v2 = u2 + 2*a*s]

0 = voy2 + 2*ay*V

Or 0 = (vo sin)2 + 2(-9.8)*V

Gives m.

c) Calculate the total horizontal distance (H) the bullet travels based on its initial angle of ascent and calculated arc of travel??

The time of flight T (total time for which the projectile remains in air) can also be calculated from the vertical motion as when the projectile reaches back to the initial horizontal level, the net vertical displacement is zero. Hence using the second equation of motion

[s = u*t + ½ a*t2]

0 = voy*T + 0.5*ay*T2

Or 0 = (vo sin)*T + 0.5*(-9.8)*T2

Gives T = 0 and s

Now as in horizontal direction, the projectile is having zero acceleration i.e. moving with uniform velocity, the horizontal distance covered by the bullet is given by

H = horizontal velocity*time taken

Or H = v0 cos *T = 500*cos 600*88.37 = 500*0.5*88.37 = 22092.5 m

Or the horizontal distance covered by the bullet is 22.0925 km.

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