Question

A ball swings counter clockwise in a vertical circle at the end of a rope 1.50 mlong. When the ball is$\mathbf{36}\mathbf{.}\mathbf{9}\mathbf{°}$past the lowest point on its way up, its total acceleration is,$\left(-22.5\hat{i}+20.2\hat{j}\right)\mathbf{ }{\mathbf{\text{m/s}}}^{\mathbf{\text{2}}}$. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.

Short answer

(a) The vector diagram showing the components of acceleration is shown in the figure

(b) The magnitude of radial acceleration is$29.7 {\text{m/s}}^{\text{2}}$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The length of rope is$1.50 \text{m}$ .
• The angle is$36.9°$
• The total acceleration is$(-22.5\hat{i}+20.2\hat{j}) {\text{m/s}}^{\text{2}}$ .

Here, the coefficient $\hat{i}$of is the magnitude of the acceleration of the x-component and the coefficient of $\hat{j}$is the magnitude of the acceleration of the -component.

The formula to calculate radial acceleration

The formula to calculate radial acceleration

${\mathit{a}}_{\mathbf{\text{r}}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

Here, vis the speed of ball and ris the radius of curvature of the length of rope. The velocity is always in the tangential direction.

Sketch a vector diagram the components of acceleration

The component of acceleration along the length of the rope due to both x and y - components is the radial acceleration.

(a)

Consider the vector diagram given below:

Determination of the radial acceleration

(b)

The component of acceleration along the length of the rope due to both x and y - components is the radial acceleration.

Write the expression for radial acceleration from the above figure

$\begin{array}{rcl}{a}_{\text{r}}& =& \left(20.2 {\text{m/s}}^{\text{2}}\right)\cos \left(36.9°\right)+\left(22.5 {\text{m/s}}^{\text{2}}\right)\cos \left(90°-36.9°\right)\\ & =& 29.7 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

Therefore, the magnitude of radial acceleration is $\begin{array}{rcl}& & 29.7 {\text{m/s}}^{\text{2}}\end{array}$.

Determination of the speed of the ball

The speed and magnitude of the velocity of the ball is $6.67 \text{m/s}$and velocity is in the direction tangential to the ball.

Write the formula to calculate radial acceleration

$a_{\text{r}}=\frac{v^2}{r}$

Here, v is the speed of ball and r is the radius of curvature of the length of rope.

The velocity is always in the tangential direction.

Substitute the above equation and we get,

$\begin{array}{rcl}29.7 {\text{m/s}}^{\text{2}}& =& \frac{v^2}{1.50 \text{m}}\\ v^2& =& \left(29.7 {\text{m/s}}^2\right)(1.50 \text{m})\\ v& =& \sqrt{\left(29.7 {\text{m/s}}^2\right)(1.50 \text{m})}\\ & =& 6.67 \text{m/s}\\ & & \end{array}$

Therefore, the speed of ball is $\begin{array}{rcl}& & 6.67 \text{m/s}\end{array}$and velocity is in the direction tangential to the ball.