You are driving down a road at 110 km/h when you notice a cop car. You slow down

Question

You are driving down a road at 110 km/h when you notice a cop car. You slow down to 85 km/h so you can get below the speed limit. (Apparently you are driving in Mexico, as the speed limits are given in km/h.) As you slow down, your tires, which have a diameter of 0.80 m, make 75 revolutions. What was the angular acceleration of the tires during this time? If you continue to slow down at the same rate, how much time passes from when you first started to slow down to when you completely stop? Two forces act on a 55-kg object. One force has magnitude 65 N directed 59 degree clockwise from the positive x-axis, and the other has a magnitude 35 N at 32 degree clockwise from the positive y-axis. Draw a free-body diagram of the object. (Assume the object is not resting on any surface.) What is the magnitude and direction of this object's acceleration? You add a third force of 42 N directed 18 degree clockwise from the negative x-axis. What is the magnitude and direction of a fourth force that would cause this object to be in equilibrium?

Explanation / Answer

2.)

a.)

angular acceleration (alpha) = change in angular velocity (omega)over time

So omegaf = vf/r = (85km/h*1000m/km*1hr/3600s)/0.40m = 23.6m/s/0.4m = 59 rad/s

omegai = vi/r = (110*1000/3600)/0.40 = 30.5/0.4 = 76.25rad/s

We find alpha using omegaf^2 = omegai^2 + 2*alpha*theta ..
So alpha = (omegaf^2 - omegai^2)/(2*theta) = (59^2 - 76.25^2)/(2*75rev*2pi rad/rev) = -2.5 rad/s^2

-2.5 rad/s^2

b) Now omegaf = 0 so using omegaf = 0 = omegai + alpha*t ....
we get t = -omegai/alpha = - 59/-2.5 = 23.6 s

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