1)A concrete highway curve of radius 70.0 m is banked at a 17.0 angle. What is t

Question

1)A concrete highway curve of radius 70.0 m is banked at a 17.0 angle.

What is the maximum speed with which a 1100 kg rubber-tired car can take this curve without sliding? (Take the static coefficient of friction of rubber on concrete to be 1.0.

2)A 140 g ball on a 60-cm-long string is swung in a vertical circle about a point 200 cmabove the floor. The tension in the string when the ball is at the very bottom of the circle is 4.0 N . At the very bottom of the circle, a very sharp knife is suddenly inserted, as shown in the figure (Figure 1) , to cut the string directly below the point of support.

How far to the right of where the string was cut does the ball hit the floor?

Explanation / Answer

1)

v = sq rt ( (tan x + u)/(1 - u * tan x) * r *g )

v = velocity
x = angle =17 degree
u = coefficient of friction = 1
r = radius = 70 m
g = 9.8 m/s^2
m = mass gets cancelled on both sides of the equation during derivation using Force-net = Force-centripetal

I get v = 35.91 m/s as the maximum speed.

2)

The force of gravity on the ball is F = mg = 0.14 x 9.8 = 1.37 newtons.
So

the tension due to the rotation is 4–1.37 = 2.62 newtons.

now

use the equations of circular motion to get the speed of rotation.

Centripetal force f = mV²/r = mr²

acceleration (outward) at rim is a = r²

where V is the tangental velocity

r is radius of circle

f = mV²/r

2.62 = (0.14)V² / 0.6m
V = 3.35 m/s

Now

you need the equations of a projectile. It is moving hoizontally with the speed above, at a height above the ground of 200–60 = 140 cm or 1.4 m

time for an object to fall 1.4 meters
t = (2d/g)
t = (2*1.4/9.8) = 0.535 seconds

Now

how far horizontally will it move in that time
3.35 m/s x 0.535 seconds = 1.79 m

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