Question

How do I answer this:

A point on a rotating turntable 20.0 cm from the center accelerates from rest to a final speed of 0.700 m/s in 1.75 s. At t = 1.25 s, find the magnitude and direction of (a) the radial acceleration, (b) the tangential acceleration,
and (c) the total acceleration of the point.

Answer 1

The magnitude of the radial acceleration is 0.4 m/s², the tangential acceleration is zero, and the total acceleration is equal to the magnitude of the radial acceleration, which is 0.4 m/s².

Step-by-step explanation:

(a) To find the magnitude of the radial acceleration, we need to calculate the change in velocity of the point over the given time interval. The final velocity is given as 0.700 m/s, and the initial velocity is zero (since the point starts from rest). The change in velocity is then 0.700 m/s - 0 m/s = 0.700 m/s. The radial acceleration can be calculated using the formula:

radial acceleration = change in velocity / time = 0.700 m/s / 1.75 s = 0.4 m/s²,

Therefore, the magnitude of the radial acceleration is 0.4 m/s².

(b) The tangential acceleration can be calculated using the formula:

tangential acceleration = radius × angular acceleration.

Since the point is rotating at a constant speed, the angular acceleration is zero, so the tangential acceleration is also zero.

(c) The total acceleration is the vector sum of the radial acceleration and the tangential acceleration. Since the tangential acceleration is zero, the total acceleration is equal to the magnitude of the radial acceleration, which is 0.4 m/s².

Answer 2

Part a)

`a_r = 1.25 m/s^2`

Part b)

`a_t = 0.4 m/s^2`

Part c)

`a_(net) = 1.31 m/s^2`

Step-by-step explanation:

initial speed of the turn table at t = 0 is given as

`v_i = 0`

after t = 1.75 s the speed is given as

`v_f = 0.700 m/s`

now the tangential acceleration is given as

`a_t = (v_f - v_i)/(t)`

`a_t = (0.700 - 0)/(1.75)`

`a_t = 0.4 m/s^2`

Part a)

Now the speed of the disc after t = 1.25 s is given as

`v_f = v_i + at`

here we will have

`v_f = 0 + (0.4)(1.25)`

`v_f = 0.5 m/s`

now radial acceleration is given as

`a_r = (v^2)/(R)`

`a_r = (0.5^2)/(0.20)`

`a_r = 1.25 m/s^2`

Part b)

tangential acceleration is given as

`a_t = 0.4 m/s^2`

Part c)

total acceleration is given as

`a_(net) = √(a_r^2 + a_t^2)`

`a_(net) = √(1.25^2 + 0.4^2)`

`a_(net) = 1.31 m/s^2`