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Kay [80]
2 years ago
5

A ball that has a mass of 0.25 kg spins in a circle at the end of a 1.6 m rope. the ball moves at a tangential speed of 12.2 m/s

. what is the centripetal force acting on the ball? 1.9 n 23 n 59 n 93 n
Physics
1 answer:
NARA [144]2 years ago
4 0

The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.

<h3>What is centripetal force?</h3>

The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.

The given data in the problem is;

m is the mass of A ball = 0.25 kg

r is the radius of circle= 1.6 m rope

v is the tangential speed = 12.2 m/s

\rm F_C is the centripetal force acting on the ball

The centripetal force is found as;

\rm F_C = \frac{mv^2}{r}  \\\\ F_C = \frac{0.25 \times (12.2)^2}{1.6}  \\\\ F_C=23.26\ N

Hence the centripetal force acting on the ball will be 23.26 N.

To learn more about the centripetal force refer to the link;

brainly.com/question/10596517

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A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

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$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

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