Impulse and Momentum on Baseball

Equation of motion for a particle of mass $m$ is $\sum F = ma = m \frac{dv}{dt}$
Rearranging terms and integrating between the limits $v = v1$ at $t = t1$ and $v = v2$ at $t = t2$ :
$\int{t1}^{t2} \sum F dt = \int{v1}^{v2} m dv = mv2 - mv1$
This is referred to as the principle of linear impulse and momentum.

$\int F dt$ is called linear impulse, and it measures the effect of a force during the time the force acts.
The impulse acts in the same direction as the force, and its magnitude has units of force-time, $\text{N} \cdot \text{s}$ .
$I = \int F dt$

The equation is rewritten as:
$mv1 + \sum \int{t1}^{t2} F dt = mv_2$
Principle of linear impulse and momentum in its $x$ , $y$ , $z$ components is:
- $m(vx)1 + \sum \int{t1}^{t2} Fx dt = m(vx)2$
- $m(vy)1 + \sum \int{t1}^{t2} Fy dt = m(vy)2$
- $m(vz)1 + \sum \int{t1}^{t2} Fz dt = m(vz)2$

Block A and B have a mass of 3 kg and 5 kg respectively. If the system is released from rest, determine the velocity of block B in 6 s.

The weight of each block is constant, and the cord tensions will also be constant.
Since the mass of pulley D is neglected, the cord tension is $TA = 2TB$ .

Block A:
$m (v1)A + \sum \int{t1}^{t2} Fy dt = m (v2)A$
$-TA (6) + 3(9.81) (6) = 3 (vA)_2$
Block B:
$m (v1)B + \sum \int{t1}^{t2} Fy dt = m (v2)B$
$TB (6) - 5(9.81) (6) = 5 (vB)_2$

We have $2sA + sB = l$
Taking the time derivative yields $2vA = -vB$ . When B moves downward, A moves upward.
Substitute this result into Eq. 1 and solve Eqs. 1 and 2 to yield $(vB)2 = 35.8 \text{ m/s}$ and $T_B = 19.2 \text{ N}$ .

Angular momentum, $H_O$ , of a particle about point O is defined as the “moment” of the particle’s linear momentum about O.

If a particle is moving along a curve, the angular momentum can be determined using a scalar formulation.
The magnitude of $HO$ is $(HO)_z = (d)(mv)$
- $d$ is the moment arm or perpendicular distance from O to the line of action of $mv$ .
Common units for $(HO)z$ are $\text{kg} \cdot \text{m}^2 \text{/s}$ .
The direction of $H_O$ is defined by the right-hand rule.

If the particle is moving along a space curve, the vector cross product is used to determine the angular momentum about O.
We have $H_O = r \times mv$
Angular momentum is determined by evaluating the determinant:
$HO = \begin{vmatrix} i & j & k \ rx & ry & rz \ mvx & mvy & mv_z \end{vmatrix}$

The moments about Point O of all forces acting are related by $\sum MO = r \times \sum F = r \times m \dot{v} = \dot{H}O$
The moments of the forces about point O can be obtained by $\frac{d}{dt} (r \times mv) = \dot{r} \times mv + r \times m \dot{v}$
Derivation of $r \times mv$ can be written as $\dot{H}_O = \frac{d}{dt} (r \times mv) = \dot{r} \times mv + r \times m \dot{v}$
Since $\sum F = m \dot{v}$ , $\dot{r} \times mv = m(\dot{r} \times \dot{r}) = 0$ .
Therefore, $\sum MO = \dot{H}O$

We have $\sum MO dt = dHO$ and integrated, assuming at time $t = t1$ , $HO = (HO)1$ and time $t = t2$ , $HO = (HO)2$
$\int{t1}^{t2} \sum MO dt = (HO)2 - (HO)1$
This equation is referred to as the principle of angular impulse and momentum
$(HO)1 + \int{t1}^{t2} \sum MO dt = (HO)2$

Since the moment of a force about point O is $MO = r \times F$ , the angular impulse may be expressed in vector form as $\int{t1}^{t2} MO dt = \int{t1}^{t2} (r \times F) dt$
The principle of angular impulse and momentum for a system of particles may be written as
$(HO)1 + \sum \int{t1}^{t2} MO dt = (HO)2$

Using impulse and momentum principles, it is possible to write equations that define the particle’s motion:
$mv1 + \sum \int{t1}^{t2} F dt = mv2$ $H{O1} + \sum \int{t1}^{t2} MO dt = HO$

If the particle is confined to move in the x-y plane, three independent equations may be written to express the motion:
$m (vx)1 + \sum \int{t1}^{t2} Fx dt = m (vx)2$
$m (vy)1 + \sum \int{t1}^{t2} Fy dt = m (vy)2$
$H{O1} + \sum \int{t1}^{t2} MO dt = H_O$

Given a 5-kg slender bar with the motion shown, we want to determine its angular momentum about point G and about the IC at this instant.
$v_A = 2 \text{ m/s}$
The bar undergoes general plane motion. The IC is established, so that
$\omega = \frac{2 \text{ m/s}}{4 \cos 30^\circ} = 0.5774 \text{ rad/s}$
$v_G = (0.5774 \text{ rad/s})(2 \text{ m}) = 1.155 \text{ m/s}$
Calculating Angular Momentum:
- About point G:
 $HG = IG \omega = \frac{1}{12} (5 \text{ kg})(4 \text{ m})^2 (0.5774 \text{ rad/s}) = 3.85 \text{ kg} \cdot \text{m}^2 \text{/s}$
- About the IC:
 $H{IC} = IG \omega + d(mv) = \left[\frac{1}{12} (5 \text{ kg})(4 \text{ m})^2\right] (0.5774 \text{ rad/s}) + (2 \text{ m})(5 \text{ kg})(1.155 \text{ m/s}) = 15.4 \text{ kg} \cdot \text{m}^2 \text{/s}$

A 10-kg disk is acted upon by a constant couple moment of 6 N-m and a force of 50 N applied to a cord wrapped around its periphery. Determine the angular velocity of the disk two seconds after starting from rest. Also, find the force components of reaction at the pin.
$I = \frac{1}{2} m r^2 = \frac{1}{2} (10 \text{ kg}) (0.2 \text{ m})^2 = 0.2 \text{ kg} \cdot \text{m}^2$
Applying Impulse and Momentum Principles:
- $\sum \int Fx dt = m ((vA)x)2$ ; $0 + A_x (2 \text{s}) = 0$
- $\sum \int Fy dt = m ((vA)y)2$ ; $0 + A_y (2 \text{s}) - [10(9.81) \text{ N}](2 \text{ s}) - 50 \text{ N}(2 \text{ s}) = 0$
- $\sum \int MA dt = I \omega2$ ; $0 + (6 \text{ N} \cdot \text{m})(2 \text{ s}) + [50 \text{ N}(2 \text{ s})](0.2 \text{ m}) = 0.2 \omega_2$
Solving:
- $A_x = 0$
- $A_y = 148 \text{ N}$
- $\omega_2 = 160 \text{ rad/s}$

A 100-kg spool with a radius of gyration $k_G = 0.35 \text{ m}$ . A cable is wrapped around the central hub, and a horizontal force $P = (t + 10) \text{ N}$ is applied, where $t$ is in seconds. If the spool is initially at rest, determine its angular velocity in 5 s. Assume the spool rolls without slipping at A.
Applying the Principle of Angular Impulse and Momentum about point A:
$\sum \int MA dt = IA \omega2$ $\int0^{5s} (t + 10) \text{ N} dt (0.75 \text{ m} + 0.4 \text{ m}) = [100 \text{ kg} (0.35 \text{ m})^2 + (100 \text{ kg})(0.75 \text{ m})^2] \omega2$ $(\frac{1}{2} t^2 + 10t) \Big|0^{5s} (1.15 \text{ m}) = 68.5 \omega_2$
Solving:
$\omega_2 = 1.05 \text{ rad/s}$