Define kinematics and dynamics. Find the angle between vectors A = 3i + 4j + 5k and B = 3i + 4j − 5k.

$$\vec{A} \cdot \vec{B} = (3)(3) + (4)(4) + (5)(-5)$$ $$\vec{A} \cdot \vec{B} = 9 + 16 - 25$$ $$\vec{A} \cdot \vec{B} = 0$$

$$|\vec{A}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$$ $$|\vec{B}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$$

$$\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$$ $$\cos \theta = \frac{0}{\sqrt{50} \cdot \sqrt{50}}$$ $$\cos \theta = 0$$

$$\theta = \cos^{-1}(0)$$ $$\theta = 90^\circ$$

The angle between the two vectors is $\boxed{90^\circ}$.

1(c) The position of a Particle moving on an x axis is given by $x=7.8+9.2t-2.1t^3$, with x in meters and t in seconds. What is its velocity at $t=3.5s$? Is the velocity constant or is it continuously changing?

$$x(t) = 7.8 + 9.2t - 2.1t^3$$ $$v(t) = \frac{dx}{dt} = \frac{d}{dt}(7.8 + 9.2t - 2.1t^3)$$ $$v(t) = 0 + 9.2 - 2.1(3t^2)$$ $$v(t) = 9.2 - 6.3t^2$$

$$v(3.5) = 9.2 - 6.3(3.5)^2$$ $$v(3.5) = 9.2 - 6.3(12.25)$$ $$v(3.5) = 9.2 - 77.175$$ $$v(3.5) = -67.975 \, \text{m/s}$$

The velocity at $t=3.5 \, \text{s}$ is $\boxed{-67.975 \, \text{m/s}}$.

The velocity function $v(t) = 9.2 - 6.3t^2$ depends on time $t$. Since $t^2$ is a variable term, the velocity changes as time progresses.
The velocity is continuously changing.

2(a) Define the following terms:

• i) Angular momentum: Angular momentum is a measure of the rotational inertia of an object in motion about a reference point or axis. For a point particle, it is the product of its moment of inertia and angular velocity.

• ii) Torque: Torque is a twisting force that tends to cause rotation about an axis. It is the rotational equivalent of linear force and is calculated as the product of force and the perpendicular distance from the pivot to the line of action of the force.

2(b) Show that the expression for rotational kinetic energy of a body is, $K=1/2 I\omega^2$. All parameters have their usual meaning.

Consider a rigid body rotating about a fixed axis with an angular velocity $\omega$. We can imagine the body as being composed of many small particles, each with mass $m_i$ at a distance $r_i$ from the axis of rotation.

$$KE_i = \frac{1}{2} m_i v_i^2$$

where $v_i$ is the tangential speed of the particle.

$$KE_i = \frac{1}{2} m_i (r_i \omega)^2$$ $$KE_i = \frac{1}{2} m_i r_i^2 \omega^2$$

$$K = \frac{1}{2} I \omega^2$$

This shows that the expression for rotational kinetic energy of a body is $K = \frac{1}{2} I \omega^2$.

2(c) Moment of inertia of a ring is $3 \, \text{Kg}m^2$. It is rotated for 20 seconds, from its rest position by a torque of $6 \, \text{Nm}$. Calculate the work done.

Given:
Moment of inertia, $I = 3 \, \text{kg} \cdot \text{m}^2$
Time, $t = 20 \, \text{s}$
Initial angular velocity, $\omega_0 = 0 \, \text{rad/s}$ (from rest)
Torque, $\tau = 6 \, \text{N} \cdot \text{m}$

$$\alpha = \frac{\tau}{I}$$ $$\alpha = \frac{6 \, \text{N} \cdot \text{m}}{3 \, \text{kg} \cdot \text{m}^2}$$ $$\alpha = 2 \, \text{rad/s}^2$$

Since the ring starts from rest ($\omega_0 = 0$):

$$\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2$$ $$\Delta \theta = (0)(20 \, \text{s}) + \frac{1}{2}(2 \, \text{rad/s}^2)(20 \, \text{s})^2$$ $$\Delta \theta = 0 + \frac{1}{2}(2)(400)$$ $$\Delta \theta = 400 \, \text{rad}$$

$$W = \tau \Delta \theta$$ $$W = (6 \, \text{N} \cdot \text{m})(400 \, \text{rad})$$ $$W = 2400 \, \text{J}$$

The work done is $\boxed{2400 \, \text{J}}$.