Respuesta :

184860

Answer:

a. At point P, tension is to the right, gravity is pulling the mass down, and the centripetal force is directed toward the circle's center, so it is likewise to the right. Tension and centripetal force are both increasing at point Q, which is at the bottom of the circle. The force of gravity is pulling you down.

b. All forces are descending at point Z. Fc = mac, therefore. T + mg = m(Vmin)2 / R Vmin2 = R(T+mg) / m Vmin = radical R(T + mg) /m c T + mg = m(Vmin)2 / R Vmin2 = R(T+mg) / m Vmin = radical R(T + mg) /m c The only difference is that the maximum tension is at pt Q. mac = fc Tmax - mg = m(Vmax)2 / R Tmax - mg = m(Vmax)2 / R

c. The same as before, but with the maximum tension at pt Q Fc = mac Tmax - mg = m(Vmax)2 / R Tmax - mg = m(Vmax)2 / R R(Tmax - mg) / m d = radical R(Tmax - mg) / m d = radical R(Tmax - mg) / m It rises at the same rate as if it were in a circle. The only force acting on it is gravity. Is this accurate? In points P, Q, and Z, I'm not clear where the centripetal force is.

Explanation:

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onyebuchinnaji

The expression for the minimum speed the ball can have at point Z (top of the circle) without leaving the circular path is [tex]v_{min} = \sqrt{\frac{r(T + W)}{m} }[/tex].

Minimum speed of an object moving in a cricular path

The minimum speed of an object moving in a circular path is calculated as follows;

Considering top of the horizontal circle;

T = Fc - mg

T = mv²/r - mg

mv²/r = T + mg

mv² = r(T + mg)

v² = r(T + mg)/m

[tex]v = \sqrt{\frac{r(T + mg)}{m} } \\\\v_{min} = \sqrt{\frac{r(T + W)}{m} }[/tex]

where;

  • W is the weight of the ball
  • m is mass of the ball
  • T is the tension in the string
  • r is the radius of the circle

Thus, the expression for the minimum speed the ball can have at point Z (top of the circle) without leaving the circular path is [tex]v_{min} = \sqrt{\frac{r(T + W)}{m} }[/tex].

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