A small steel sphere of mass 0.40 kg is attached by a 0.50 m long cord to a swivel pin set into the surface of a frictionless table top. The sphere moves in a circle on the horizontal surface with a speed of 6.0 m/s.
a.) what is the magnitude of the centripetal acceleration of the sphere?
b.) what is the tension of the cord?
To find the magnitude of the centripetal acceleration of the sphere, we can use the formula:
\[a_{\text{c}} = \frac{v^2}{r}\]
where \(a_{\text{c}}\) is the centripetal acceleration, \(v\) is the velocity, and \(r\) is the radius of the circle.
a.) To calculate the magnitude of the centripetal acceleration, substitute the given values into the formula:
\[a_{\text{c}} = \frac{(6.0 \, \text{m/s})^2}{0.50 \, \text{m}}\]
Simplifying this expression:
\[a_{\text{c}} = \frac{36.0 \, \text{m}^2/\text{s}^2}{0.50 \, \text{m}}\]
\[a_{\text{c}} = 72.0 \, \text{m/s}^2\]
So, the magnitude of the centripetal acceleration of the sphere is \(72.0 \, \text{m/s}^2\).
b.) To find the tension of the cord, we need to consider the forces acting on the sphere. The tension in the cord provides the centripetal force that keeps the sphere moving in a circle. This centripetal force is given by:
\[F_{\text{c}} = \frac{mv^2}{r}\]
where \(F_{\text{c}}\) is the centripetal force, \(m\) is the mass of the sphere, \(v\) is the velocity, and \(r\) is the radius of the circle.
The tension in the cord is equal to the centripetal force, so:
\[F_{\text{c}} = T\]
Substituting the given values into the formula:
\[T = \frac{mv^2}{r}\]
\[T = \frac{(0.40 \, \text{kg})(6.0 \, \text{m/s})^2}{0.50 \, \text{m}}\]
Simplifying this expression:
\[T = \frac{14.4 \, \text{kg} \, \text{m}^2/\text{s}^2}{0.50 \, \text{m}}\]
\[T = 28.8 \, \text{N}\]
Therefore, the tension in the cord is \(28.8 \, \text{N}\).