Question

The angular velocity of a flywheel obeys the equation ωz(t)=A+Bt2, where t is in seconds and A and B are constants having numerical values 2.65 (for A) and 1.60 (for B).

Required:
a. What are the units of A and B if ωz is in rad/s?
b. What is the angular acceleration of the wheel at (i) t = 0 and (ii) t = 5.00 s?
c. Through what angle does the flywheel turn during the first 2.00 s?

Answer 1

Answer:

Explanation:

a )

Iz( t ) = A + B t²

Iz( t ) = angular velocity

putting dimensional formula

T⁻¹ = A + Bt²

A = T⁻¹

unit of A is rad s⁻¹

BT² = T⁻¹

B = T⁻³

unit of B is rad s⁻³

b )

Iz( t ) = A + B t²

dIz / dt = 2Bt

angular acceleration = 2Bt

at t = 0

angular acceleration = 0

at t = 5

angular acceleration = 2 x 1.6 x 5

= 16 rad / s²

Iz( t ) = A + B t²

dθ / dt = A + B t²

integrating ,

θ = At + B t³ / 3

when t = 0 , θ = 0

when t = 2

θ = At + B t³ / 3

= 2.65 x 2 + 1.6 x 2³ / 3

= 5.3 + 4.27

= 9.57 rad .

Flywheel turns by 9.57 rad during first 2 s .

Answer 2

(a) The unit of A is rad/s and the unit of B is rad/s³.

(b) The angular acceleration of the flywheel at 0 s is 0 and at 5 s is 16 rad/s²

(c) The angular displacement of the flywheel during the first 2 seconds, is 9.57 rad.

The given parameters;

• z(t) = A + Bt²
• A = 2.65 and B = 1.6

The units of A and B if z(t) is in radian per second (rad/s), are calculated as follows;

$z(t)[\frac{rad}{s} ] = A[\frac{rad}{s} ] \ + \ Bt^2[\frac{rad}{s^3} ]$

Thus, the unit of A is rad/s and the unit of B is rad/s³.

The angular acceleration of the flywheel is calculated as follows;

$a = \frac{d\omega }{dt} =2Bt$

when, t = 0

a = 2(1.6)(0) = 0

when t = 5 s

a = 2(1.6)(5) = 16 rad/s²

The angular displacement of the flywheel during the first 2 seconds, is calculated as follows;

$\theta = \int\limits {z(t)} \, dt\\\\\theta = At \ + \ \frac{Bt^3}{3} \\\\when, \ t = 2\ s;\\\\\theta = (2.65\times 2) \ + \ (\frac{1.6\times 2^3}{3} )\\\\\theta = 9.57 \ rad$

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