An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
1 answer:
Answer:
1) A
2) C
3) B
4) A
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8) A
9) C
10) C
Explanation:
1) m = 15kg, a = , F = ma = 15*10 = 150N
2) m = 3kg, v = , r = 4m, F =
= 12N
3) a = , r = 10m, v=?
F = and F = ma
equating the two equations and cancelling a, we have:
= a
making v the subject of formula, we have:
v =
=
= 10
4) r = 10m, v = , a = ?
F =
F = ma
equating the above equations and making a subject of formula, we get:
a =
a = 25/10 = 2.5
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8) F =
assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:
F =
Now, if v is tripled
F =
F = 9
We can see that the force will be 9X greater than it was.
9) F =
assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:
F =
Now, if v is doubled
F =
F = 4
We can see that the force will be 4X greater than it was.
10) F =
assuming m and v is unity, that is the values are 1 respectively
F = 1/r
if r is doubled,
F = 1/2 * 1/r
We can see that the force is 1/2 as big as it was
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Answer:
a) u_s=0.375
b )v=14.4 m/s
Explanation:
1 Concepts and Principles
Particle in Equilibrium: If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:
∑F=0 (1)
2- Particle in Uniform Circular Motion:
If a particle moves in a circle or a circular arc of radius Rat constant speed v, the particle is said to be in uniform circular motion. It then experiences a net centripetal force F and a centripetal acceleration a_c. The magnitude of this force is:
F=ma_c
=m*v^2/R (2)
where m is the mass of the particle F and a_c are directed toward the center of curvature of the particle's path.
<em>3- The magnitude of the static frictional force between a static object and a surface is given by:</em>
f_s=u_s*n (3)
where u_s is the coefficient of kinetic friction between the object and the surface and n is the magnitude of the normal force.
Given Data
R (radius of the car's path) = 170 m
v (speed of the car) = 25 m/s
Required Data
- In part (a), we are asked to find the coefficient of static friction u_s that will prevent the car from sliding.
- In part (b), we are asked to find the speed of the car v if the coefficient of static friction is one-third u_s found in part (a).
Solution:
see the attachment pic
Since the car is not accelerating vertically, we can model it as a particle in equilibrium in the vertical direction and apply Equation (1)
∑F_y=n-mg
= mg (4)
We model the car as a particle in uniform circular motion in the horizontal direction. The force that enables the car to remain in its circular path is the force of static friction at the point of contact between road and tires. Apply Equation (2) to the horizontal direction:
∑F_x=f_s
=m*v^2/R
Substitute for f_s from Equation (3):
u_s=m*v^2/R
Substitute for n from Equation (4):
u_s*mg=m*v^2/R
u_s*g = v^2/R
Solve for u_s:
u_s= v^2/Rg (5)
Substitute numerical values:
u_s=0.375
(b)
The new coefficient of static friction between the tires and the pavement is:
u_s'=u_s/3
Substitute u_s/3 for u_s in Equation (5):
u_s/3=v^2/Rg
Solve for v:
v=14.4 m/s
