Centripetal force and gravitation

meech-ward commented 7 years ago

Learning Material

Distributed Practice

Varsity Tutors

Distributed Practice questions answered correctly:

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[x] 5 page 2 #1

Extra:

[x] 1 page 2 #2
[x] 2 page 3 #1Example Question #21

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Correct: 3
Incorrect: 0

Quiz

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meech-ward commented 7 years ago

Today I spent about an hour watching video, taking notes, and practicing centripetal acceleration.

centripetal acceleration = (magnitude of velocity squared) / (radius of the circle)

meech-ward commented 7 years ago

Today I spent about 2 hours working on Centripetal force. I find that there's a lot to think about when dealing with Centripetal force, and I'm mainly trying to remember to only calculate forces that are in the same direction. I understand it, I'm just slow with it.

Next time I'm on to Gravity.

meech-ward commented 7 years ago

Today I spent about 40 minutes on gravity videos and 30 minutes on the quiz.

Now I'm on to Work and Energy

meech-ward commented 7 years ago

Example Question #21 : Centripetal Force And Acceleration

On a wet road, the coefficient of static friction between a car's tires and the flat road is $\mu_s=0.4$ $\displaystyle \mu_s=0.4$ . What is the maximum speed a car can safely navigate a turn with a 30m $\displaystyle 30m$ radius of curvature?

Possible Answers:

$117.7 \frac{m}{s}$ $\displaystyle 117.7 \frac{m}{s}$

$3.46 \frac{m}{s}$ $\displaystyle 3.46 \frac{m}{s}$

$12.0 \frac{m}{s}$ $\displaystyle 12.0 \frac{m}{s}$

$26.0 \frac{m}{s}$ $\displaystyle 26.0 \frac{m}{s}$

$10.85 \frac{m}{s}$ $\displaystyle 10.85 \frac{m}{s}$

Correct answer:

$10.85 \frac{m}{s}$ $\displaystyle 10.85 \frac{m}{s}$

Explanation:

When a car is going around a flat turn, the only force keeping the car going along a circular path is the force of friction, which is equal to the centripetal force. We do not need the mass of the car to calculate the maximum speed, as we will show below.

F_c=F_f $\displaystyle F_c=F_f$

$\frac{mv^2}{r}=\mu_sN$ $\displaystyle \frac{mv^2}{r}=\mu_sN$

N=mg $\displaystyle N=mg$

Rearrange these equations and solve for the velocity.

$v = \sqrt{\mu_s g r}$ $\displaystyle v = \sqrt{\mu_s g r}$

$v= \sqrt{0.4(9.81\frac{m}{s^2})30m}=10.85\frac{m}{s}$ $\displaystyle v= \sqrt{0.4(9.81\frac{m}{s^2})30m}=10.85\frac{m}{s}$

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