# A bowling ball weighing 71.2 N is attached to the ceiling by a 3.80 m rope. The ball is pulled to one side and released; it then swings back and forth like a pedulum. as the rope swings through its lowest point, the speed of the bowling ball is measured at 4.20 m/s. At that instant, find a) the magnitude and direction of the acceleration of the bowling ball and b) the tension in the rope.

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1. a) The magnitude of the acceleration is V^2/R and it is directed along the rope (i.e. vertical), perpendicular to the direction of motion.

b) If the rope tension is T, Newton's Second Law tells you that

F = T - Mg = Ma = M V^2/R

Solve for T

T = M*(g + V*2/R)

R is the rope length.

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2. So for a I got 17.64 / 3.80 = 4.64 sec is this correct

and then for b i got 71.2( 9.8+ 4.20)*2 / 3.80 and got 524.63

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3. LMBO!!!

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4. thanks

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5. Since it's a pendulum, we we'll use the formula for circular acceleration.
Therefore; a=v^/r
a=v^2/r
=(4.2)^2/3.8
=4.64m/s^2
Since the circular acceleration is always directed to the center so the acceleration of the bowling ball is directed to the center.

b) To find the tension, first we have to find the mass using F=ma.
F=ma
71.2=m(9.81)
m=7.23kg
We cannot use the acceleration to be 9.81, instead we'll use (9.81- 4.64)
Weight of the swinging pendulum=Force=ma=(7.23)(9.81- 4.64)=37.5N

*Now we’ll use Newton’s second law;
T-mg=ma
T-(7.23)(9.81)=(7.23)(9.81-4.46)
T-71.2=(7.23)(5.17)
T-71.2=37.5
T=37.5+71.2
T=108.7N

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6. There's two of my answers above so chose whichever you prefer

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8. hell naw

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9. Since it's a pendulum, we we'll use the formula for circular acceleration.
Therefore; a=v^/r
a=v^2/r
=(4.2)^2/3.8
=4.64m/s^2
Since the circular acceleration is always directed to the center so the acceleration of the bowling ball is directed to the center.

b) To find the tension, first we have to find the mass using F=ma.
F=ma
71.2=m(9.81)
m=7.23kg
We cannot use the acceleration to be 9.81, instead we'll use (9.81- 4.64)
Tension=Force=ma=(7.23)(9.81- 4.64)=37.5N

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