Question

The axis of a circular cylinder coincides with the z-axis and it rotates with an angular velocity of (3i−5j +9k) rad/s. Determine i. The tangential velocity at a point P on the cylinder, whose co-ordinates are ( j +5k) metres,

Answer 1

Answer:

The tangential velocity at point P is $\vec v_{P} = -34\,\hat{i} -15\,\hat{j}+3\,\hat{k}\,\,\left[\frac{m}{s} \right]$.

Explanation:

Vectorially speaking, we define tangential velocity at point P ($\vec v_{P}$), measured in meters per second, by the following vectorial expression:

$\vec v_{P} = \vec \omega \times \vec r_{P}$ (1)

Where:

$\vec \omega$ - Angular velocity of the cylinder, measured in radians per second.

$\vec r_{P}$ - Radius of rotation, measured in meters.

If we know that $\vec \omega = 3\,\hat{i}-5\,\hat{j}+9\,\hat{k}\,\left[\frac{rad}{s} \right]$ and $\vec r_{P} = \hat{j}+5\,\hat{k}\,[m]$, then the tangential velocity at point P is:

$\vec v_{P} = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\3\,\frac{rad}{s} &-5\,\frac{rad}{s} &9\,\frac{rad}{s} \\0\,m&1\,m&5\,m\end{array}\right|$

$\vec v_{P} = \left[\left(-5\,\frac{rad}{s} \right)\cdot (5\,m)-(1\,m)\cdot \left(9\,\frac{rad}{s} \right)\right]\,\hat{i}-(5\,m)\cdot \left(3\,\frac{rad}{s}\right)\,j +\left(3\,\frac{rad}{s} \right)\cdot (1\,m)\,\hat{k}$

$\vec v_{P} = -34\,\hat{i} -15\,\hat{j}+3\,\hat{k}\,\,\left[\frac{m}{s} \right]$

The tangential velocity at point P is $\vec v_{P} = -34\,\hat{i} -15\,\hat{j}+3\,\hat{k}\,\,\left[\frac{m}{s} \right]$.