Question

During the spin cycle the time dependent angular speed of a washing machine drum is given by the equation ω(t) = at + bt2 - ct4 where a = 2.6 rad/s2, b = 0.85 rad/s3 and c = 0.035 rad/s5. At time t = 0 s, a point P on the washer drum is located at θ0 = 0.75 rad. Write an equation for the angular acceleration of the drum, as a function of time, in terms of the given parameters.

Answer 1

Answer:

$a(t)=a+2bt-4ct^{3}$

Explanation:

We have the following angular speed equation :

$w(t)=at+bt^{2}-ct^{4}$

We also know the parameters :

$a=2.6\frac{rad}{s^{2}}$

$b=0.85\frac{rad}{s^{3}}$

$c=0.035\frac{rad}{s^{5}}$

We need the equation of the angular acceleration in terms of a,b and c.

We know that the angular acceleration is the rate of change of angular speed respect time.

Therefore, to obtain the equation a(t) we need to derivate w(t) in respect of  the variable t.

$w(t)=at+bt^{2}-ct^{4}$ ⇒ derivating in respect of t ⇒

$a(t)=a+2bt-4ct^{3}$

And that is the angular acceleration equation in terms of a,b and c.

Answer 2

Answer:

α = 2.6 +1.7 t - 0.14 t³

Explanation:

Given that

ω(t) = at + bt² - ct⁴

where a = 2.6 rad/s², b = 0.85 rad/s³ and c = 0.035 rad/s⁵

We  know that angular acceleration is the rate of change of angular velocity

α = dω/dt

ω(t) = at + bt² - ct⁴

dω/dt= a + 2 b t - 4 ct³

So

α = a + 2 b t - 4 ct³

By putting the values of a b and c

α = a + 2 b t - 4 ct³

α = 2.6 + 2 x 0.85 t - 4 x 0.035 t³

α = 2.6 +1.7 t - 0.14 t³