What is the tension, the force causing centripetal acceleration, and speed?

Question

An air puck of mass 0.32 kg is tied to a string and allowed to revolve in a circle of radius 1.3 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of 0.9 kg is tied to it. The suspended mass remains in equilibrium while the puck on the tabletop revolves.
(a) What is the tension in the string?
(b.)What is the force causing the centripetal acceleration on the puck?
(c)What is the speed of the puck?

Answer 1

In summary:

(a) Tension in string = (0.32kg)*g = 3.1392 N (if g=9.81 m/s/s)

(b) The centripetal acceleration is caused by the tension in the string, which is (0.32kg)*g = 3.1392 N (if g=9.81 m/s/s)

(c) The speed of the puck is sqrt(g*(1.3m)) = 2.74703 m/s (if g=9.81 m/s/s).

To explain:

You know that the puck of mass (0.32kg) is being pulled down to earth with a force of (0.32kg)*g, where g is the acceleration due to gravity. However, it is not moving, so it's net acceleration is zero. That means that the tension in the string must be EQUAL to the WEIGHT of the puck. So . . .

(a) Tension in string = (0.32kg)*g = 3.1392 N (if g=9.81 m/s/s)

This tension in the string is the same force pulling the puck in at all times. That is, the puck keeps trying to move in a straight line, but the tension in the string keeps pulling it toward the hole. Thus, instead of moving in a straight line, the puck moves in a circle. This "centripetal" (meaning "toward the center") force is the tension in the string. So the answer to (b) is . . .

(b) The centripetal acceleration is caused by the tension in the string, which is (0.32kg)*g = 3.1392 N (if g=9.81 m/s/s)

Finally, since we know the puck is traveling in a CIRCLE, then we know that the force on it must be equal to m*v*v/r, where m is the mass of the puck, v is the speed of the puck, and r is the radius of the circle. This is the expression for centripetal FORCE (see the source below for explanation of this formula). However, we know the centripetal force is equal to (0.32kg)*g. So all we do is set (0.32kg)*g equal to m*v*v/r and solve for v, the speed of the puck. Note that m = 0.32kg and r=1.3m, so we have:

0.32kg*g = (0.32kg)*v*v/(1.3m)

Multiply by sides by (1.3m) and divide both sides by (0.32kg) to get:

g*(1.3m) = v*v

Now take the square root of both sides to get:

sqrt(g*(1.3m)) = v

If we assume g=9.81m/s/s, then we have:

sqrt( (9.81 m/s/s)*(1.3m) ) = v

which is:

2.74703 m/s = v

So the answer to (c) is:

(c) The speed of the puck is sqrt(g*(1.3m)) = 2.74703 m/s (if g=9.81 m/s/s).


So, again, in summary:

(a) Tension in string = (0.32kg)*g = 3.1392 N (if g=9.81 m/s/s)

(b) The centripetal acceleration is caused by the tension in the string, which is (0.32kg)*g = 3.1392 N (if g=9.81 m/s/s)

(c) The speed of the puck is sqrt(g*(1.3m)) = 2.74703 m/s (if g=9.81 m/s/s).

Answer 2

it is a count number of framing somewhat. in case you sit down in the reference body of helicopters fan, you'll locate that the pressure is amazingly by using centrifugal rigidity. As centrifugal/ centripetal rigidity = mv^2/R the position m is mass of the blade