Centripetal and Tangential Acceleration

In summary, a car is driving on a circular road with radius R, and the distance driven is given by d(t) = at^3 + bt. The tangential and radial acceleration in terms of a, b, and R are found by taking the second and first derivatives of d(t) respectively. The tangential acceleration is 6a when t = 2 seconds, and the radial acceleration is (3a(2)^2 + b)^2 / R. Despite a(t) not being zero, this approach is correct because the car is not moving at uniform circular motion.
  • #1
Arkun
4
0

Homework Statement


A car drives on a circular road with radius ##R##. The distance driven by the car is given by ##d(t) = at^3 + bt## [where ##t## in seconds will give ##d## in meters]. In terms of ##a##, ##b##, and ##R##, and when ##t = 2## seconds, find an expression for the magnitudes of (i) the tangential acceleration ##a_{tan}##, and (ii) the radial acceleration ##a_{rad}##.

Homework Equations


$$d(t) = at^3 + bt$$
$$a_{rad} = \frac{v^2}{r}$$
$$a_{tan} = \frac{d|\vec v|}{dt}$$

The Attempt at a Solution


For i) I took the second derivative of ##d(t)## and plugged in ##t = 2## to get:

$$a_{tan} = 6a(2)$$

My reasoning for this is that ##d(t)## describes the distance the car traveled around circular road, which means that its derivative will describe the car's tangential velocity at time ##t##, and ultimately means that its second derivative will be the tangential acceleration.

For ii) I took the first derivative of ##d(t)## because, like with part i), I reasoned it must be the tangential velocity of the car since ##d(t)## describes the distance driven by the car along the curved road. So I first took the derivative of ##d(t)## and got:

$$v(t) = 3a(2)^2 + b$$

Since the formula for ##a_{rad}## is quotient of the square of the tangential velocity and the radius of the circle, I plugged ##v(t)## into ##v## and ##R## into ##r## for the formula of ##a_{rad}## to get:

$$a_{rad} = \frac{(3a(2)^2 + b)^2}{R}$$

The issue I'm having with this problem is that I don't meet the conditions stated in the problem for part i): I don't have ##a_{tan}## in terms of ##a##,##b##, and ##R##. So what I'm left wondering is whether ##v(t)## actually is the tangential velocity or if I need to use another formula that will help me meet the conditions stated in the problem. Since ##a(t)## is not zero I'm assuming the car isn't moving at uniform circular motion, so uniform circular motion equations don't apply.
 
Physics news on Phys.org
  • #2
Arkun said:
The issue I'm having with this problem is that I don't meet the conditions stated in the problem for part i): I don't have ##a_{tan}## in terms of ##a##,##b##, and ##R##. So what I'm left wondering is whether ##v(t)## actually is the tangential velocity or if I need to use another formula that will help me meet the conditions stated in the problem. Since ##a(t)## is not zero I'm assuming the car isn't moving at uniform circular motion, so uniform circular motion equations don't apply.
The time derivative of the distance traveled is speed, a scalar, not 'tangential velocity' which is a vector. But you need speed to get the radial acceleration. The time derivative of speed is the tangential acceleration. So your derivation is correct. The problem asked the tangential and radial acceleration in terms of a, b and R. You did it. If a variable does not appear in a formula, you can consider that it is with zero coefficient somewhere. If d(t) was (at^3+bt)^2, for example, both a and b would appear in the tangential acceleration.
 

FAQ: Centripetal and Tangential Acceleration

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and its magnitude is equal to v^2/r, where v is the velocity of the object and r is the radius of the circle.

2. How is tangential acceleration different from centripetal acceleration?

Tangential acceleration is the acceleration experienced by an object along its tangent direction, perpendicular to the centripetal acceleration. It is caused by a change in the magnitude or direction of the object's velocity. Unlike centripetal acceleration, tangential acceleration does not contribute to the object's circular motion, but rather affects its speed or direction.

3. What is the relationship between tangential and centripetal acceleration?

Tangential acceleration and centripetal acceleration are both components of the total acceleration experienced by an object in circular motion. The total acceleration is the vector sum of the two components, and its direction is always changing as the object moves along the circular path.

4. How is centripetal acceleration related to centripetal force?

According to Newton's Second Law, an object experiencing a net force will accelerate in the direction of that force. In the case of circular motion, the centripetal force is the net force acting on the object, causing it to accelerate towards the center of the circle. Therefore, centripetal acceleration is directly related to the centripetal force acting on the object.

5. What are some real-life examples of centripetal and tangential acceleration?

Centripetal acceleration can be observed in many everyday activities such as riding a bike around a curve, swinging a ball on a string, or driving a car around a roundabout. Tangential acceleration is commonly seen in activities like accelerating a car or changing the direction of a projectile's motion, such as a basketball player shooting a hoop.

Back
Top