User:Prof McCarthy/Work

Work by gravity

The force of gravity between on a mass m exerted by another mass M is given by

$\mathbf {F} =-{\frac {GmM}{r^{3}}}\mathbf {r} ,$

where r is to position vector from M to m.

Let the mass m move at the velocity v then the work of gravity on this mass as it moves from position r(t₁) to r(t₂) is given by

$W=-\int _{\mathbf {r} (t_{1})}^{\mathbf {r} (t_{2})}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot d\mathbf {r} =-\int _{t_{1}}^{t_{2}}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot {\dot {\mathbf {r} }}dt.$

Notice that the position and velocity of the mass m are given by

$\mathbf {r} =r\mathbf {e} _{r},\qquad {\dot {\mathbf {r} }}={\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t},$

where e_r and e_t are the radial and tangential unit vectors directed relative to the vector from M to m. Use this to simplify the formula for work of gravity to,

$W=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}(r\mathbf {e} _{r})\cdot ({\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t})dt=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}r{\dot {r}}dt={\frac {GMm}{r(t_{2})}}-{\frac {GMm}{r(t_{1})}}.$

This calculation uses the fact that

${\frac {d}{dt}}r^{-1}=-r^{-2}{\dot {r}}=-{\frac {\dot {r}}{r^{2}}}.$

The function

$V={\frac {GMm}{r}},$

is known as the gravitational potential function, also know as gravitational potential energy.

gravitational potential Φ(r) at a point in gravitational field around mass M, such as a planet, is defined as the work done in bringing a unit mass from infinity to that point. The force of gravity on a mass m is directed along the line connecting the two bodies, M and m,

$F=-{\frac {GMm}{r^{2}}},$

where G is the gravitational constant and r is the distance between the two bodies.

The work of this force on the particle m from infinity to the distance r is given by

$\Phi (r)=-\int _{\infty }^{r}{\frac {GMm}{r^{2}}}dr={\frac {GMm}{r}}{\bigg |}_{\infty }^{r}={\frac {GMm}{r}},$

because dividing by infinity yields zero.

Work by gravity K11-K12 version

The gravitational potential Φ(r) at a point in gravitational field around mass M, such as a planet, is defined as the work done in bringing a unit mass from infinity to that point. The force of gravity on a mass m is directed along the line connecting the two bodies, M and m,

$F=-{\frac {GMm}{r^{2}}},$

where G is the gravitational constant and r is the distance between the two bodies.

The work of this force on the particle m from infinity to the distance r is given by

$\Phi (r)=-\int _{\infty }^{r}{\frac {GMm}{r^{2}}}dr={\frac {GMm}{r}}{\bigg |}_{\infty }^{r}={\frac {GMm}{r}},$

because dividing by infinity yields zero.

Vehicle coasting down a mountain road

Consider the case of a vehicle that starts at rest and coasts down a mountain road, the work-energy principle helps compute the minimum distance that the vehicle travels to reach a velocity of say 60mph (88 fps). Rolling resistance and air drag will slow the vehicle down so the actual distance will be less, than if these forces are neglected.

Let the trajectory of the vehicle down the road be X(t). The driving force on the vehicle is the constant force of gravity F=(0,0,W), while the constraint forces R of the road on the vehicle vary along the trajectory. Newton's second law yields,

$\mathbf {F} +\mathbf {R} =m{\ddot {\mathbf {X} }}.$

The scalar product of this equations with the velocity, V=(v_x, v_y,v_z), yields

$Wv_{z}=m{\dot {V}}V,$

where V is the magnitude of V. Integrate of both sides to obtain

$\int _{t_{1}}^{t_{2}}Wv_{z}dt={\frac {m}{2}}V^{2}(t_{2})-{\frac {m}{2}}V^{2}(t_{1}).$

Recall that V(t_1)=0, and because W is constant along the trajectory, the integral of the vertical velocity is the vertical distance, therefore

$W\Delta z={\frac {m}{2}}V^{2}.$

Notice that this result does not depend the shape of the road followed by the vehicle.

In order to determine the distance along the road, assume the downgrade is 6%, which is a steep road. This means the altitude decreases 6 feet for every 100 feet traveled--for angles this small the sin and tan functions are approximately equal. Therefore, the distance s in feet traveled along the road is

$s={\frac {\Delta z}{0.06}}=8.3gV^{2}.$

Notice that this formula uses the fact that the mass of the vehicle is m=W/g.

Work of forces on a rigid body

The work of forces acting at various points on a single rigid body can be calculated from the work of a resultant force and torque. For example, let the forces F₁, F₂ ... F_n act on the points X₁, X₂ ... X_n in a rigid body.

The trajectories of X_i, i=1,...,n are defined by the movement of the rigid body. This movement is given by the set of rotations [A(t)] and the trajectory d(t) of a reference point in the body. Let the coordinates x_i i=1,...,n define these points the moving rigid body's reference frame M, so that the the trajectories traced in the fixed frame F are given by

$\mathbf {X} _{i}(t)=[A(t)]\mathbf {x} _{i}+\mathbf {d} (t)\quad i=1,\ldots ,n.$

The velocity of the points X_i along their trajectories are

$\mathbf {V} _{i}={\vec {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+{\dot {\mathbf {d} }},$

where ω is the angular velocity vector obtained from the skew symmetric matrix

$[\Omega ]={\dot {A}}A^{T},$

known as the angular velocity matrix.

The small amount of work by the forces over the instant of time δt is given by

$\delta W=\mathbf {F} _{1}\cdot \mathbf {V} _{1}\delta t+\mathbf {F} _{2}\cdot \mathbf {V} _{2}\delta t+\ldots +\mathbf {F} _{n}\cdot \mathbf {V} _{n}\delta t$

or

$\delta W=\sum _{i}^{n}\mathbf {F} _{i}\cdot ({\vec {\omega }}\times (\mathbf {X} _{i}-\mathbf {d} )+{\dot {\mathbf {d} }})\delta t.$

This formula can be rewritten to obtain

$\delta W=(\sum _{i}^{n}\mathbf {F} _{i})\cdot {\dot {\mathbf {d} }}\delta t+(\sum _{i}^{2}(\mathbf {X} _{i}-\mathbf {d} )\times \mathbf {F} _{i})\cdot {\vec {\omega }}\delta t=\mathbf {F} \cdot {\dot {\mathbf {d} }}\delta t+\mathbf {T} \cdot {\vec {\omega }}\delta t,$

where F and T are the resultant force and torque applied at the reference point d of the moving frame M in the rigid body.