Question

Winter Work Packet 2018 Th is packet is designed to help you study on kinematics and rotational motion during the break. It is requested that you complete this packet if you currently have less than a 65%, but all students are welcome to complete it. This packet will be due the first day of school, I will not accept it any later. I will be available to help with any questions regarding the problems via REMIND on Wednesday the second from 2PM to 5PM Section 1: Definitions and conversions 1. Perform the following conversions a. 30° to radians b. 2t radians to degrees c. 4 revolutions to radians d. 210° to radians When dealing with a rolling ball, which equation will tell us the linear distance the ball has travelled if we know the angle which it has covered? Show that by dividing by time, the conversion between tangential distance and 2. 3. ular distance becomes the conversion between tangential velocity and ang angular velocity What will happen to the following values if I were to double my radius: 4. a. Tangential velocity b. Angular velocity c. Centripetal force d. Moment of inertia e. Torque 5. Identify the following variables and which angular value they correspond to: a. b. C, θ e. r Section 2: Rotational Calculations

Answer 1

1. To perform the following conversions which are given below as -

(a) 30⁰ to radians

converting deg into rad :

$\theta$ = [30⁰ x ($\pi$ / 180⁰)] rad

$\theta$ = ($\pi$ / 6) rad

$\theta$ = 0.523 rad

(b) 2$\pi$ radians to degree

converting rad to deg :

$\theta$ = [2$\pi$ x (180⁰ / $\pi$)] deg

$\theta$ = 360⁰

(c) 4 revolutions to radians

converting rev to rad :

$\theta$ = [4 x (2$\pi$)] rad

$\theta$ = (8$\pi$) rad

$\theta$ = 25.1 rad

(d) 210⁰ to radians

converting deg into rad :

$\theta$ = [210⁰ x ($\pi$ / 180⁰)] rad

$\theta$ = (7$\pi$ / 6) rad

$\theta$ = 3.66 rad

5. To identify the following variables which are given below as -

(a) I $\Rightarrow$ moment of inertia

(b) $\omega$$\Rightarrow$ angular velocity

(c) $\theta$$\Rightarrow$ angular distance

(d) $\alpha$$\Rightarrow$ angular acceleration

(e) $\tau$$\Rightarrow$ torque

Which angular value they correspond to :

$\tau$ = I $\alpha$

6. (a) The moment of inertia of a ball which will be given by -

I = (2/5) m r²

I = [(0.4) (2 kg) (0.7 m)²]

I = 0.392 kg.m²

(b) The applied torque which will be given by -

$\tau$ = r F sin $\theta$

$\tau$ = (0.7 m) (5 N) sin 80⁰

$\tau$ = 3.44 N.m

(c) The angular acceleration which will be given by -

$\alpha$ = $\tau$ / I

$\alpha$ = [(3.44 N.m) / (0.392 kg.m²)]

$\alpha$ = 8.77 rad/s²

8. (a) Using equation of rotational motion (1), we have

$\omega$_f = $\omega$_i + $\alpha$ t

where, $\omega$_i = initial angular velocity of a flywheel = 4 rev/s = 25.1 rad/s

$\omega$_f = final angular velocity of a flywheel = 8 rev/s = 50.2 rad/s

$\alpha$ = angular acceleration = $\pi$/3 rad/s² = 1.04 rad/s²

then, we get

t = [(50.2 rad/s) - (25.1 rad/s)] / (1.04 rad/s²)

t = [(25.1 rad/s) / (1.04 rad/s²)]

t = 24.1 sec

(b) Using equation of rotational motion (3), we have

$\omega$_f² = $\omega$_i² + 2 $\alpha$$\theta$

(50.2 rad/s)² = (25.1 rad/s)² + 2 (1.04 rad/s²) $\theta$

[(2520.04 rad²/s²) - (630.01 rad²/s²)] = (2.08 rad/s²) $\theta$

$\theta$ = [(1890.03 rad²/s²) / (2.08 rad/s²)]

$\theta$ = 908.6 rad