Question

A small object with mass 4.00 kg moves counter-clockwise with constant angular speed 1.50 rad/s in a circle of radius 3.00 m centered at the origin. It starts at the point with position vector $3.00\hat{i}m$. It then undergoes an angular displacement of 9.00 rad. (a) What is its new position vector? Use unit-vector notation for all vector answers. (b) In what quadrant is the particle located, and what angle does its position vector make with the positive x axis? (c) What is its velocity? (d) In what direction is it moving? (e) What is its acceleration? (f) Make a sketch of its position, velocity, and acceleration vectors. (g) What total force is exerted on the object?

Short answer

The new position vector of object is $\left(-2.74\hat{i}+1.22\hat{j}\right)m$.

Step-by-step solution

Identification of given data

The mass of object is m = 4 kg

The constant angular speed of object is $\omega =1.50rad/s$

The radius of circle is R = 3m

The angular displacement of object is $\delta =9rad$

Conceptual Explanation

The angular speed of an object is the rate of variation of the angular displacement of the object with time.

Determination of new position vector

The angle for the new position vector is given as:

$$\begin{gathered} \alpha =\delta \left(\frac{180°}{\pi }\right)-360 \\ =\left(9rad\right)\left(\frac{180°}{\pi rad}\right)-360° \\ =156° \end{gathered}$$

The new position vector of object is given as:

$$\begin{gathered} \overrightarrow{r}=R\left(\cos \alpha \hat{i}+\sin \alpha \hat{j}\right) \\ =\left(3m\right)\left[\left(\cos 156°\right)\hat{i}+\left(\sin 156°\right)\hat{j}\right] \\ =\left(-2.74\hat{i}+1.22\hat{j}\right)m \end{gathered}$$

Therefore, the new position vector of object is $\left(-2.74\hat{i}+1.22\hat{j}\right)m$.