A car with mass 1600 kg drives around a flat circular track of radius 28.0 m. The coefficient of friction between the car tires
1 answer:
Answer:
v=15.24 m/s
Explanation:
Given that
Mass ,m= 1600 kg
radius ,r= 28 m
Coefficient of friction ,μ = 0.83
The radial force on the car when it takes turn
The friction force on the car
Fr= μ m g
The condition for motion without losing traction
F= Fr
v²=μ r g
Now by putting the values
( take g=10m/s²)
v=15.24 m/s
The speed of the car will be 15.24 m/s
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Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m