Question

A couple of problems I was struggling with on my previous assignment

3. A fire dancer in china town makes circles with a ball on a string that is 25cm long. The amount of rotations the fireball makes count it to be 4 times every second.

a. Find the speed and acceleration of the fireball.

The dancer lets go of the string and the ball shoots into the air straight at you with an angle of 35 degrees to the ground. Assume when the ball is let go it is 1.5 m from the ground.

b. You are standing 6 m away from the fireball. Does it hit you and make you catch on fire? Why or why not?

c. What is the speed of the ball when it hits the ground?

8. 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.

a. What was the angular acceleration of the tires during this time?

b. 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?

16. Two forces act on a 55-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other has a magnitude 35 N at 32° clockwise from the positive y-axis.

a. Draw a free-body diagram of the object. (Assume the object is not resting on any surface.)

b. What is the magnitude and direction of this object’s acceleration?

c. You add a third force of 42 N directed 18° 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?

Answer 1

a) Time period of revolution of the ball is T = 1/4 = 0.25 s 4 Angular velocity w = 2 pi/T = 6.28/0.25 = 25.12 rad/s Tangential Speed of the ball v = rw = 0.25 times 25.12 = 6.28 m/s Radial acceleration of the ball a = w^2 r = 25.12 times 0.25 = 157.8 rad/s^2 b) Time flight of the ball when it hits the ground, Position along the vertical y = 1.5 + v_y t + 4.9t^2 When y = 0: 1.5 + 6.28 times sin 35 times t - 4.9 t^2 = 0 t = 1.03 s (Only positive solution is considered) Horizontal displacement of the ball in this time x = v cos 35 times T = 6.28 times 0.819 times 1.03 = 529 m Since you are standing 6 m away, you will not catch fire by the ball. Speed of the ball when it hits the ground, Horizontal component of the velocity