Radial and tangential acceleration

[steph35]

Homework Statement

A point on a rotating turntable 21.5 cm from the center accelerates from rest to a final speed of 0.800 m/s in 1.75 s. At t = 1.28 s, find the magnitude and direction of each of the following.

Homework Equations

(a) the radial acceleration

(b) the tangential acceleration

(c) the total acceleration of the point
? m/s^2, ? °

The Attempt at a Solution

for this i used the right equations for the give times and substituded V= 2pir/T but i didnt get the right answer. i used a_r= -v^2/r and a_t= d|v|/dt...can somebody explain it in a simple way...

 

[alphysicist]

Hi steph35,

Homework Statement

A point on a rotating turntable 21.5 cm from the center accelerates from rest to a final speed of 0.800 m/s in 1.75 s. At t = 1.28 s, find the magnitude and direction of each of the following.

Homework Equations

(a) the radial acceleration

(b) the tangential acceleration

(c) the total acceleration of the point
? m/s^2, ? °

The Attempt at a Solution

for this i used the right equations for the give times and substituded V= 2pir/T but i didnt get the right answer. i used a_r= -v^2/r and a_t= d|v|/dt...can somebody explain it in a simple way...

What numbers did you use, and what answers did you get for the three accelerations?

 

[baba89]

I understand your confusion and will provide a detailed explanation of the concept of radial and tangential acceleration.

First, let's define what radial and tangential acceleration are. Radial acceleration is the acceleration of an object moving in a circular path towards the center of the circle. On the other hand, tangential acceleration is the acceleration of an object moving in a circular path tangent to the circle.

Now, let's apply this concept to the given scenario. We have a point on a rotating turntable that is 21.5 cm from the center. This point accelerates from rest to a final speed of 0.800 m/s in 1.75 s. We are asked to find the magnitude and direction of the radial and tangential acceleration at t = 1.28 s.

To solve for the radial acceleration, we can use the equation a_r = v^2/r, where v is the velocity and r is the distance from the center. In this case, the velocity at t = 1.28 s is 0.800 m/s, and the distance from the center is 21.5 cm or 0.215 m. Plugging these values into the equation, we get a_r = (0.800 m/s)^2 / 0.215 m = 2.963 m/s^2.

For the tangential acceleration, we can use the equation a_t = dv/dt, where v is the velocity and t is the time. Since we are given the final speed of 0.800 m/s, we can use this as the velocity. However, we need to find the change in velocity, or dv, from t = 0 to t = 1.28 s. We can do this by dividing the change in speed by the change in time, which is (0.800 m/s - 0 m/s) / (1.28 s - 0 s) = 0.625 m/s^2.

Now, to find the total acceleration, we can use the Pythagorean theorem. Since the radial and tangential accelerations are perpendicular to each other, we can treat them as the sides of a right triangle. The total acceleration would then be the hypotenuse of this triangle, which we can solve using the equation a_tot = √(a_r^2 + a_t^2). Plugging in the values, we get