Answer to Question #321609 in Math for Pheebs

Question

A point particle is travelling in a circular path with variable speed. If its

velocity at time t seconds is v = (2t + 1)θˆ:

a. What is the tangential component of acceleration at any time t ?

b. What is the centripetal acceleration at any time t ?

c. What the acceleration of the particle at any time t? Your answer should

look like a = acentriˆr + atan ˆθ, where acentri and atan are scalars which

represent the centripetal and tangential components, respectively, of

the total acceleration a.

Answer 1

The position of a particle can be expressed as

$$\vec{r}=r\cos \theta \hat{i}+r\sin \theta \hat{j}$$

The unit vector $\cos \theta \hat{i}+\sin \theta \hat{j}$ can be written as $\hat{r}.$

This vector is not constant since it changes over time. When computing the time derivatives of the position we must take into account the relations

$$\dfrac{d\hat{r}}{dt}=\dot{\theta}\hat{\theta}$$

$$\dfrac{d\hat{\theta}}{dt}=-\dot{\theta}\hat{r}$$

Given $v(t)=(2t+1)\hat{\theta}$

Taking the derivative of the velocity with respect to time

$$\dfrac{d\vec{v}}{dt}=2\hat{\theta}-(2t+1)\dot{\theta}\hat{r}$$

The angular velocity can be expressed in terms of the radius as

$$\dot{\theta}=v/R=(2t+1)/R$$

Substitute

$$\vec{a}(t)=\dfrac{d\vec{v}}{dt}=2\hat{\theta}-\dfrac{(2t+1)^2}{R}\hat{r}$$

a. The first term of the acceleration is the tangential acceleration

$$a_t=2$$

b. The centripetal acceleration is

$$a_r=\dfrac{(2t+1)^2}{R}$$

Its minus sign indicates that it points toward the center of the circular path. To evaluate the centripetal acceleration, we must know the value of the radius.

c.

The total acceleration is given by the expression

$$\vec{a}(t)=2\hat{\theta}-\dfrac{(2t+1)^2}{R}\hat{r}$$

$$|\vec{a}(t)|=\sqrt{(2)^2+(\dfrac{(2t+1)^2}{R})^2}$$