Question

A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 = 4ft/s. Knowing that the tape does not slip on the drums, determine (a) the angular acceleration of drum B, and (b) the number of revolutions executed by drum B during the 4-s interval. 5) A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13000 ft/s. Knowing that the rocket is fired vertically from the ground, determine its acceleration (a) as it is fired, and (b) as the last particle of fuel is being consumed.

Answer 1

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

$v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2$

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

$v = \omega R\\a = \alpha R$

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

$v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft$

One revolution is equal to the circumference of the drum. So, total number of revolutions is

$x / (2\pi R) = 6/(\pi R)$

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

$a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2$

b) $a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2$