Question

A bowling ball weighing 71.2 N is attached to the ceiling by a 3.60 m rope. The ball is pulled to one side and released; it then swings back and forth like a pendulum. As the rope swings through its lowest point, the speed of the bowling ball is measured at 3.90 m/s.

A. at that instant, find the magnitude of the acceleration of the bowling ball.
B. At that instant, find the direction of the acceleration of the bowling ball.
upward/downward/in the direction of motion
C. At that instant, find the tension in the rope.

Answer 1

Answer:

a)$a=4.22\ m/s^2$

b)upwards

c)T=101.83 N

Explanation:

Given that

Weight of ball = 71.2 N

Mass ,m=Wt/g

Mass of ball ,m=7.25 kg

r= 3.6 m

V= 3.9 m/s

a)

The acceleration of the ball a

$a=\dfrac{V^2}{r}$

$a=\dfrac{3.9^2}{3.6}$

$a=4.22\ m/s^2$

b)

The direction of the acceleration will be upwards.

c)

To find the tension in the rope

$T-mg=\dfrac{mV^2}{r}$

$T=mg+\dfrac{mV^2}{r}$

$T=71.2+\dfrac{7.25\times 3.9^2}{3.6}$

T=101.83 N

Answer 2

a) The acceleration is 4.22 m/s²

b) The direction of acceleration is upwards

c) The tension in the rope is 101.83 N

Given that the weight of the ball is 71.2 N is so its mass will be :

m = 71.2/9.8 = 7.25 kg

Radius r = 3.6 m

velocity of the ball v = 3.9 m/s

Circular motion:

(a) In circular motion the acceleration of the ball is given by:

$a=\frac{v^2}{r}\\\\a=\frac{3.9\times3.9}{3.6}\\\\a=4.22\;m/s^2$

(b) the direction of acceleration in circular motion is towards the center, so in this case, it will be upwards.

(c) The tension in the rope:

The equation of motion can be written as:

$T-mg=\frac{mv^2}{r}\\\\T=mg+m\frac{v^2}{r}\\\\T=71.2N+7.25\times4.22N\\\\T=101.83N$

Learn more about circular motion:

https://brainly.com/question/2285236?referrer=searchResults