Problem statement

Problem 1.69. A bar of mass m resting on a smooth horizontal plane starts moving due to the force F = mg/3 of constant magnitude. In the process of its rectilinear motion, the angle α between the direction of this force and the horizontal varies as α = as, where a is a constant, and s is the distance traversed by the bar from its initial position. Find the velocity of the bar as a function of the angle α.

We assume a = 1 per metre to start, so the angle turns by 1 radian (57.3°) for every metre travelled. The problem uses a letter only. It is a slider.

We assume g = 9.8 m/s² and m = 1 kg. The mass only sets the forces in newtons, and the speed does not depend on it. Both are sliders.

We assume the angle is measured in radians in α = a s, so a has units of 1/m.

Symbols: s is the distance travelled, v the speed, F the pulling force of fixed size mg/3, and N the normal force.

The big idea in plain words

The force always has the same size, but it turns as the bar moves. At the start it is horizontal and pushes the bar hard. As the angle grows, more of the force lifts and less of it pushes forward. At 90° it points straight up and no longer speeds the bar up. After that it points backwards and slows the bar.

Because the push depends on where the bar is, and not on the time, the quickest tool is energy: the speed at any place comes from the work done up to that place.

The solution in each frame

Only the ground frame is used. The plane is fixed to the ground, so this frame is inertial and Newton's second law holds as written. No second frame is needed.

The solution step by step

Step 1, the equation along the plane. The plane is smooth, so only the horizontal part of F moves the bar: m dv/dt = F cos α = (mg/3) cos(a s).

Step 2, remove the time. The push depends on the distance s, so write dv/dt = (dv/ds)(ds/dt) = v dv/ds. Then m v dv = (mg/3) cos(a s) ds.

Step 3, integrate. The bar starts from rest at s = 0. On the left, the integral of v dv is v²/2. On the right, the integral of cos(a s) ds is sin(a s)/a. So v²/2 = (g/3) sin α/a.

Step 4, the speed. v = √(2g sin α/(3a)). At α = ° this is m/s.

Step 5, what the result says. The sine is largest at 90°, so the highest speed is √(2g/(3a)) = m/s. At α = 180°, which is s = π/a = m, the speed is zero again. The force then points backwards and the bar turns back. So the bar moves back and forth over a track of length π/a.

Step 6, the bar stays on the plane. The normal force is N = mg − F sin α = mg(1 − sin α/3). It is never below 2mg/3, so the bar does not lift off. Now N = N.

Step 7, a link to the pendulum. Put θ = a s − π/2. Then s'' = (g/3) cos(a s) becomes θ'' = −(a g/3) sin θ, which is the pendulum equation with amplitude 90°. Its round trip time is T = 4K/√(a g/3), with K = 1.8541 for this amplitude. Here T = s.

Five methods

1. Force method

Newton's second law along the plane, with v dv/ds in place of dv/dt. This is steps 1 to 4.

2. Energy method

The work of the force is the integral of F cos α ds from 0 to s, which equals the kinetic energy ½ m v². The integral is (mg/3) sin α/a. This is the same calculation, and it is the shortest, because no time appears.

3. Graph method

Plot F cos α against s. The area under the curve up to s is the work. For a cosine this area is the sine, which gives the same result and shows why the speed rises up to 90° and falls after it.

4. Calculus method

Solve s'' = (g/3) cos(a s). With the substitution θ = a s − π/2 it becomes the pendulum equation, whose first integral is θ'²/2 − (a g/3) cos θ = constant. It leads to the same speed and also gives the round trip time.

5. Check method

Units: g/a has the units of m²/s², so v is in m/s. Limits: for a very large a, the force turns quickly and the bar stays near the start, so v is small. For a very small a, the force stays almost horizontal over a long track and the bar reaches a high speed √(2g/(3a)). At α small, v ≈ √(2gα/(3a)) = √(2gs/3), which is the speed under a steady push of g/3 over the distance s. This agrees.

Fastest: the energy method. Most reliable: the energy method followed by the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Angle α	Yes	The position of the body (s = α/a), the force arrow, v, N and the dots on the graphs. Play then runs the real motion from there
Turning rate a	Yes	The track length π/a, v at every angle, the highest speed, the round trip time and all curves
Mass m	No	Only the forces in newtons. Not the speed
Gravity g	Yes	v, the highest speed and the round trip time. F stays one third of the weight
Given-answer mode	Sets α	The page finds the angle on the way out for a chosen speed, and warns if the speed is above the highest speed
Play and time bar	No	Play runs the real motion back and forth, holds 1.2 s at the highest speed and 1.2 s at the stop. The bar sets the angle

Parameters

Quantity	Symbol	Value used
Turning rate of the force	a	per m
Mass of the bar	m	kg
Gravity	g	m/s²
Pulling force	F = mg/3	N
Angle now, distance	α, s	°, m
Speed now	v	m/s
Highest speed, track length	vmax, π/a	m/s, m

Exam traps

• Treating the acceleration as constant. The push F cos α changes with the distance, so v² = 2 w s does not apply.
• Integrating with respect to time. The data are given as a function of the distance, so use v dv = w ds.
• Using degrees in α = a s with a in 1/m. The angle has to be in radians for the integral of cos(a s) to give sin(a s)/a.
• Adding friction. The plane is smooth, so there is none, and only F cos α acts along the plane.
• Believing the speed keeps growing. After 90° the force has a backward part, so the bar slows down and stops at 180°.