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Question 1

Curvilinear Motion — Summary of Formulas

Curvilinear motion refers to the motion of a particle along a curved path. At any point on the path, the motion can be described using tangential and normal components, involving angular speed, angular displacement, and centripetal acceleration.

The following are the fundamental formulas used to describe circular and curvilinear motion.

Variable Definitions

r → radius of circular path (meters)
s → arc length (meters)
θ → angular displacement (radians)
ω → angular speed (rad/s)
v → linear speed (m/s)
a_c → centripetal (normal) acceleration (m/s²)
T → period — time for one revolution (seconds)
f → frequency (Hz)
m → mass of object (kg)
F_c → centripetal force (N)

1. Angular Kinematics

$$ \omega = \frac{\theta}{t} $$

$$ s = r \theta $$

2. Linear and Angular Velocity Relationship

$$ v = \omega r $$

3. Centripetal (Normal) Acceleration

$$ a_c = \frac{v^2}{r} $$

$$ a_c = \omega^2 r $$

$$ a_c = \frac{\omega^2 r^2}{r} $$

4. Centripetal Force

$$ F_c = m a_c $$

$$ F_c = m \frac{v^2}{r} $$

$$ F_c = m \omega^2 r $$

5. Period and Frequency

$$ T = \frac{1}{f} $$

$$ f = \frac{1}{T} $$

$$ f = \frac{\text{cycles}}{\text{second}} $$

Frequency is measured in Hz (Hertz), where 1 Hz = 1 cycle per second.

Answer

1. Velocity Vector (first derivatives)

x-component:

$$ x(t) = 3t^3 - 2t $$ $$ \dot{x}(t) = 9t^2 - 2 $$ $$ \dot{x}(2) = 34 $$

y-component:

$$ y(t) = -4t^{1/2} - t $$ $$ \dot{y}(t) = -2t^{-1/2} - 1 $$ $$ \dot{y}(2) = -2\left(\frac{1}{\sqrt{2}}\right) - 1 \approx -2.41 $$

z-component:

$$ z(t) = 3t^2 - 2 $$ $$ \dot{z}(t) = 6t $$ $$ \dot{z}(2) = 12 $$

Magnitude of velocity:

$$ v = \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{z}^2} $$ $$ v = \sqrt{34^2 + (-2.41)^2 + 12^2} $$ $$ v \approx 36.14\ \text{m/s} $$

2. Acceleration Vector (second derivatives)

x-component:

$$ \ddot{x}(t) = 18t $$ $$ \ddot{x}(2) = 36 $$

y-component:

$$ \ddot{y}(t) = -2\left(-\tfrac{1}{2}\right)t^{-3/2} $$ $$ \ddot{y}(2) = \frac{1}{\sqrt{8}} \approx 0.35 $$

z-component:

$$ \ddot{z}(t) = 6 $$

Magnitude of acceleration:

$$ a = \sqrt{\ddot{x}^2 + \ddot{y}^2 + \ddot{z}^2} $$ $$ a = \sqrt{36^2 + (0.35)^2 + 6^2} $$ $$ a \approx 36.50\ \text{m/s}^2 $$

Final Answers:

Velocity magnitude: $36.14\ \text{m/s}$
Acceleration magnitude: $36.50\ \text{m/s}^2$

Dynamics of Rigid Bodies – Problem: – Diagram

Answer

1. Integrate the velocity to obtain position.

$$ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} $$ $$ d\mathbf{r} = \mathbf{v}(t)\,dt $$ $$ \mathbf{r}(t) = \int_0^t \left[ 3\mathbf{i} + (6 - 2t)\mathbf{j} \right] dt $$

Integrate components:

$$ r_x(t) = 3t $$ $$ r_y(t) = 6t - t^2 $$ $$ \mathbf{r}(t) = (3t)\mathbf{i} + (6t - t^2)\mathbf{j} $$

2. Evaluate position at the limits.

At $t = 1$:

$$ \mathbf{r}(1) = 3\mathbf{i} + 5\mathbf{j} $$

At $t = 3$:

$$ \mathbf{r}(3) = 9\mathbf{i} + 9\mathbf{j} $$

3. Displacement from $t = 1$ to $t = 3$:

$$ \Delta \mathbf{r} = \mathbf{r}(3) - \mathbf{r}(1) = (9\mathbf{i} + 9\mathbf{j}) - (3\mathbf{i} + 5\mathbf{j}) $$ $$ \Delta \mathbf{r} = 6\mathbf{i} + 4\mathbf{j}\;\text{m} $$

Final Answer: The displacement is $6\mathbf{i} + 4\mathbf{j}\;\text{m}$.

Answer

1. Integrate $v_y$ to obtain $y(t)$.

$$ v_y = \frac{dy}{dt} = c t^2 $$ $$ y(t) = \int c t^2 \, dt = \frac{c t^3}{3} $$

2. Use the path relation $ y = b x^2 $ to solve for $ x(t) $.

$$ b x^2 = \frac{c t^3}{3} $$ $$ x(t) = t^{3/2} \sqrt{\frac{c}{3b}} $$

3. Compute the velocity component $v_x = dx/dt$.

$$ v_x = \frac{d}{dt} \left( t^{3/2} \sqrt{\frac{c}{3b}} \right) $$ $$ v_x = \frac{3}{2} t^{1/2} \sqrt{\frac{c}{3b}} $$

4. Compute the x-component of acceleration $a_x = dv_x/dt$.

$$ a_x = \frac{d}{dt} \left( \frac{3}{2} t^{1/2} \sqrt{\frac{c}{3b}} \right) $$ $$ a_x = \frac{3}{4} t^{-1/2} \sqrt{\frac{c}{3b}} $$

5. Compute the y-component of acceleration.

$$ a_y = \frac{dv_y}{dt} $$ $$ a_y = \frac{d}{dt}(c t^2) = 2 c t $$

Final Answers:

$$ a_x = \frac{3}{4\sqrt{t}} \sqrt{\frac{c}{3b}}, \qquad a_y = 2 c t $$

Answer

Step 1: Compute acceleration by differentiating velocity.

$$ \mathbf{a} = \frac{d\mathbf{v}}{dt} $$ $$ a_x = 32t, \qquad a_y = 12 t^2, \qquad a_z = 5 $$

At $t = 2$:

$$ a_x = 32(2) = 64 $$ $$ a_y = 12 (2^2) = 48 $$ $$ a_z = 5 $$

Acceleration magnitude:

$$ a = \sqrt{ a_x^2 + a_y^2 + a_z^2 } $$ $$ a = \sqrt{ 64^2 + 48^2 + 5^2 } $$ $$ a \approx 80.156 \ \text{m/s}^2 $$

Step 2: Integrate velocity components to get position.

$$ x = \int 16 t^2 \, dt = \frac{16 t^3}{3} $$ $$ y = \int 4 t^3 \, dt = t^4 $$ $$ z = \int (5t + 2)\, dt = \frac{5 t^2}{2} + 2 t $$

Evaluate at $t = 2$:

$$ x(2) = \frac{16(2^3)}{3} = \frac{128}{3} \approx 42.67 $$ $$ y(2) = 2^4 = 16 $$ $$ z(2) = \frac{5(2^2)}{2} + 2(2) = 10 + 4 = 14 $$

Final coordinate position:

$$ (x, y, z) = (42.67,\ 16,\ 14) $$

Answer

1. Integrate $v_y$ to obtain $y(t)$.

$$ v_y = \frac{dy}{dt} = 2r \cos 2t $$ $$ y(t) = \int 2r \cos 2t \, dt = r \sin 2t $$

2. Use the circular path equation to find $x(t)$.

$$ x^2 + y^2 = r^2 $$ $$ x^2 = r^2 - (r \sin 2t)^2 = r^2 \cos^2 2t $$ $$ x = \pm r \cos 2t $$

3. Compute $v_x$.

$$ v_x = \frac{dx}{dt} $$ $$ v_x = \frac{d}{dt} \left( \pm r \cos 2t \right) $$ $$ v_x = \pm 2r \sin 2t $$

4. Compute the acceleration components.

$$ a_y = \frac{dv_y}{dt} $$ $$ a_y = \frac{d}{dt}(2r \cos 2t) = -4r \sin 2t $$

$$ a_x = \frac{dv_x}{dt} $$ $$ a_x = \frac{d}{dt} (\pm 2r \sin 2t) $$ $$ a_x = \pm 4r \cos 2t $$

Final Answers:

$$ a_x = \pm 4 r \cos 2t, \qquad a_y = -4 r \sin 2t $$

Answer

1. Differentiate $y = 0.05 x^2$ with respect to time to find $v_y$.

$$ y = 0.05 x^2 $$ $$ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} $$ $$ v_y = (2)(0.05)x \, v_x = 0.1 x v_x $$

At $x = 5$ and $v_x = -3$:

$$ v_y = 0.1(5)(-3) = -1.5 \, \text{m/s} $$

2. Differentiate $v_y$ to get $a_y$.

Since $v_y = 0.1 x v_x$, use the product rule:

$$ a_y = \frac{dv_y}{dt} $$ $$ a_y = 0.1 \left( x \, a_x + v_x^2 \right) $$

Evaluate at $x = 5$, $v_x = -3$, $a_x = -1.5$:

$$ a_y = 0.1\left( 5(-1.5) + (-3)^2 \right) $$ $$ a_y = 0.1(-7.5 + 9) $$ $$ a_y = 0.15 \, \text{m/s}^2 $$

Final Answers:

$$ v_y = -1.5 \, \text{m/s}, \qquad a_y = 0.15 \, \text{m/s}^2 $$

Question 2

Problem:

A bob of weight W is moving with a constant velocity of 2.4m/s and is suspended on a 0.6m string. If the weight of the bob is 450N, compute the required inclination, θ, Tension T in the string, and the time required for one complete revolution.

Question 3

Curvilinear Motion in Vectors

Question 4

Problem:

The position of a particle is given by:

$$ \mathbf{r}(t) = (3t^3 - 2t)\,\mathbf{i} - (4t^{1/2} + t)\,\mathbf{j} + (3t^2 - 2)\,\mathbf{k} $$

where $t$ is in seconds. Determine the magnitude of the velocity and the magnitude of the acceleration when $t = 2$ s.

Question 5

Problem:

Problem:

The velocity of a particle is

$$ \mathbf{v}(t) = 3\mathbf{i} + (6 - 2t)\mathbf{j} \;\text{m/s} $$

If $\mathbf{r} = 0$ when $t = 0$, determine the displacement of the particle during the time interval from $t = 1\;\text{s}$ to $t = 3\;\text{s}$.

Question 6

Problem:

A particle travels along the parabolic path

$$ y = b x^2 $$

If its component of velocity along the y-axis is

$$ v_y = c t^2 $$

determine the x- and y-components of the particle's acceleration. Here, b and c are constants.

Question 7

Problem:

The velocity of a particle is given by $$ \mathbf{v} = \{ 16 t^2 \} \mathbf{i} + \{ 4 t^3 \} \mathbf{j} + \{ 5t + 2 \} \mathbf{k} \ \text{m/s}, $$ where $t$ is in seconds. If the particle is at the origin when $t = 0$, determine the magnitude of the particle's acceleration when $t = 2$ s. Also, find the $x$, $y$, and $z$ coordinate position of the particle at this instant.

Question 8

Problem:

A particle travels along the circular path $ x^2 + y^2 = r^2. $ If the y-component of the particle's velocity is $ v_y = 2r \cos 2t, $ determine the $x$ and $y$ components of its acceleration at any instant.

Question 9

Problem:

The box slides down the slope described by the equation $$ y = 0.05 x^2 $$ Dynamics of Rigid Bodies – Problem: – Diagram where $x$ is in meters. If the box has $x$ components of velocity and acceleration of $v_x = -3 \, \text{m/s}$ and $a_x = -1.5 \, \text{m/s}^2$ at $x = 5$ m, determine the $y$ components of the velocity and acceleration of the box at this instant.

Question 10

Problem: Van Travels Over the Hill

The van travels over the hill described by $y=[-1.5x10^{-3}x^2+15]ft$. If it has a constant speed of 75ft/s, determine the x and y components of the van's velocity and acceleration when x=50ft.

Question 11

Problem: Rotated Drum

A 60kg drum shown has a radius of gyration of 0.25m. A cord of negligible mass is wrapped around the periphery of the drum and attached to the block having a mass of 20kg. How long will the block reach a distance of 4 meters starting from rest, if the velocity is given by $v={8s^2-2s)m/s$, where is is in meters?

Question 12

Problem: Radius of Curvature of the Path of a Stone Thrown at an Angle

A stone is thrown with an initial velocity of 100ft per second upward at 60° to the horizontal. Compute the radius of curvature of its path at the point where it is 50ft horizontally from its initial position.

Question 13

Problem: Radius of Curvature of the Path of a Stone Thrown Horizontally

A stone is thrown with an initial velocity of 100ft per second with the horizontal. Compute the radius of curvature of its path at the point where it is 50ft horizontally from its initial position.