Question

A 2.0 kg stone is tied to a 0.50 m string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is:_____a. 0b. 6.0Nc. 8Nd. 10N

Answer 1

Answer:

The magnitude of the torque about the origin is 108 N-m.

Explanation:

Given that,

Mass of the stone, m = 2 kg

Length of the string, l = 0.5 m

Angular velocity of the stone, $\omega=12\ rad/s$

The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin, r = 0.75 m

Firstly it is required to find the centripetal force acting on the stone. It is given by :

$F=mr\omega^2$

$F=2\times 0.5\times (12)^2$

F = 144 N

Now the magnitude of the torque about the origin is given by :

$\tau=r\times F$

$\tau=0.75\ m\times 144\ N$

$\tau=108\ N-m$

So, the magnitude of the torque about the origin is 108 N-m. Hence, this is the required solution.

Answer 2

The magnitude of the torque will be "108 N/m".

Given that:

• Mass = 2.0 kg
• Length of string = 0.5 m
• Angular velocity = 12 rad/s
• Distance from origin = 0.75 m

As we know the formula,

→ $w = \frac{vt}{r}$

or,

→ $vt = w\times r$

      $= 12\times 0.5$

      $= 6$

Now,

→ $Fc = \frac{mv^2}{r}$

        $= \frac{2\times 36}{0.5}$

        $= 144$

hence,

The torque will be:

= $F\times r$

By substituting the values, we get

= $144\times 0.75$

= $108 \ N/m$

Thus the response above is appropriate.

Learn more about torque here:

https://brainly.com/question/17077402