Topic 3 of 3
Choose the radial force
Centripetal force means the inward resultant required for circular motion. Identify the real interactions first, then resolve them towards the centre.
A body moving uniformly around a circle has inward acceleration a = v2/r = rω2. Newton's second law in the inward direction gives:
= mv2/r = mrω2
In a free-body diagram, include the actual forces on the selected body: for example, tension, weight, friction or a normal contact force. Do not add a separate centripetal-force arrow to these interactions. It names the resultant's role.
Choose inward as positive for the radial equation and deal separately with other directions. In the laboratory frame used here, no outward force is needed to balance the inward resultant: the body is accelerating.
Worked horizontal tether
Tension supplies the horizontal resultant
A 0.200 kg body moves at 2.00 m/s in a circle of radius 0.500 m on a smooth horizontal table. A light taut horizontal string pulls it towards the centre. Neglect resistance and take local g = 9.81 N/kg.
The string supplies the inward horizontal force
A 0.200 kg body travels at 2.00 m/s on a smooth horizontal table, with a taut string of radius 0.500 m. Resistance is negligible; g = 9.81 N/kg.
Top view: tension before release; tangent path after removal
The purple arrow applies before removal. Removing the string removes this tension; then the dashed tangent path applies. No extra outward or inward force is added.
Side view: only the vertical balance is shown
Vertical acceleration is zero. While the string remains taut, the horizontal resultant is still 1.60 N inward, giving horizontal acceleration 8.00 m/s2.
There is no vertical acceleration, so support N balances weight mg. The only horizontal force is the string tension S:
S = mv2/r
= 0.200(2.00)2/0.500 = 1.60 N
The tension itself supplies the 1.60 N inward resultant. An additional 1.60 N centripetal force would count the same effect twice.
If the string is removed and no horizontal force remains, the body initially continues along the tangent in the table's plane. It does not depart radially outward or keep following the circle without a horizontal force.
State what remains fixed
At the same mass and radius, doubling speed from 2.00 to 4.00 m/s makes the required force 6.40 N, four times as large. By contrast:
- At fixed mass and speed, doubling radius halves mv2/r.
- At fixed mass and angular speed, doubling radius doubles mrω2.
Both conclusions are correct under their stated conditions. Speed and angular speed are connected by v = rω; they cannot both stay fixed when radius changes.
Optional check A 0.200 kg body moves on a smooth horizontal table, attached to a light horizontal string of radius 0.500 m. The string can provide at most 3.60 N tension. What is the maximum speed in this model?
Worked conical pendulum
Only one component of tension is inward
A 0.200 kg bob on a 1.00 m light inextensible string travels uniformly around a horizontal circle of radius 0.600 m. The fixed pivot is 0.800 m above the orbit plane. The string is taut and air resistance is negligible. For this example, use g = 10.0 N/kg.
Let θ be the string's angle to the vertical. The right triangle gives sin θ = 0.600/1.00 and cos θ = 0.800/1.00, so θ = 36.8699°. Name tension S to distinguish it from period T.
The circle's radius is not the string length
A 0.200 kg bob moves uniformly on a taut 1.00 m string. This example uses g = 10.0 N/kg. Use S for tension, reserving T for the period.
Side view: a 0.600 / 0.800 / 1.00 m triangle
Two real forces; tension is resolved into components
Purple arrows are the two real forces on the bob. Brown dashed arrows are components of that same tension, not additional forces.
The inward resultant is 1.50 N, giving acceleration 7.50 m/s2. Using r = 0.600 m gives speed 2.12 m/s; neither the full 2.50 N tension nor the 1.00 m string length is substituted as the radial quantity.
The bob stays at the same height, so the vertical component of tension balances its weight:
0.800S = 0.200 × 10.0 = 2.00 N
S = 2.50 N
The horizontal component points towards the centre of the horizontal orbit:
= 2.50 × 0.600 = 1.50 N
mv2/r = 1.50
Use the orbit radius 0.600 m, not the 1.00 m string length:
v = √(1.50 × 0.600/0.200)
= √4.50 = 2.12 m/s
ω = v/r = 3.54 rad/s
T = 2π/ω = 1.78 s
The full tension is 2.50 N, but the inward resultant is only 1.50 N. Its 2.00 N upward component balances the 2.00 N downward weight. Keep extra digits until the final result when moving from speed to angular speed and period.
Optional check A 0.200 kg conical-pendulum bob moves in a horizontal circle. Its tension is 2.50 N at 36.87 degrees to the vertical; take g = 10.0 N/kg. What supplies the inward resultant?
Choose a graph that tests the relationship
For the smooth-table model with fixed m = 0.200 kg and r = 0.500 m, the inward force is F = (m/r)v2. The following values are calculated model data, not force-sensor observations.
| v / (m/s) | v2 / (m2/s2) | F / N |
|---|---|---|
| 1.00 | 1.00 | 0.400 |
| 1.50 | 2.25 | 0.900 |
| 2.00 | 4.00 | 1.60 |
| 2.50 | 6.25 | 2.50 |
F plotted vertically against v horizontally is curved. Plotting F against v2 instead gives a straight line through the origin, with gradient:
Gradient units: N/(m2/s2) = kg/m
The transformed horizontal quantity is squared speed, not speed with a relabelled axis. A straight line with this gradient is consistent with the supplied model.
Use logarithms to find an exponent
Take reference values v0 = 1.00 m/s and F0 = 0.400 N from the first row. The ratios v/v0 and F/F0 have no units, so their logarithms are meaningful.
ln(F/F0) = 2 ln(v/v0)
A plot of ln(F/F0) vertically against ln(v/v0) horizontally therefore has gradient 2. At v = 2.00 m/s, the force ratio is 4, giving ln 4 / ln 2 = 2. Both ratios equal 1 at the reference row, so that transformed point is (0, 0).
More generally, a supplied power law F/F0 = (v/v0)n becomes a straight line of gradient n on these logarithmic axes. Using log10 consistently on both axes gives the same exponent. Logarithms here require positive inputs; do not take a logarithm of a zero or negative force ratio, or of an unnormalised quantity carrying units.
Worked logarithms of combined changes
Change mass, speed and radius together
Starting from m0 = 0.200 kg, v0 = 1.00 m/s and r0 = 0.500 m, the inward force is F0 = 0.400 N. In a second circular-motion model, triple the mass, double the speed and increase the radius by a factor of 1.5. Each force uses F = mv2/r.
= 3 × 22/1.5 = 8
F = 8 × 0.400 = 3.20 N
To separate these contributions using logarithms, a product becomes a sum and a quotient becomes a difference. For positive dimensionless a and b:
ln(a/b) = ln a - ln b
ln(an) = n ln a
Apply all three rules to the force ratio:
+ 2 ln(v/v0) - ln(r/r0)
= ln 3 + 2 ln 2 - ln 1.5
= 2.07944
Exponentiating recovers F/F0 = exp(2.07944) ≈ 8. Keep the unrounded logarithm for a precise calculation. Adding the logarithms accounts for multiplying the positive ratios; ln(a + b) is not ln a + ln b.
Read an orders-of-magnitude scale
lg means log10, whereas ln uses base e. The product, quotient and power rules work with either base when used consistently. On a graph of lg(F/F0), consider these positive force ratios:
| F/F0 | lg(F/F0) |
|---|---|
| 0.001 | -3 |
| 0.01 | -2 |
| 0.1 | -1 |
| 1 | 0 |
Each step of 1 on this logarithmic axis means multiplying the force ratio by 10. Moving from -3 to 0 spans three orders of magnitude: the ratio increases from 0.001 to 1, a factor of 103 = 1000, not an increase of 3 N. Such a scale displays multiplicative comparisons across a wide range; it cannot include zero or negative ratios.
What a real force comparison needs
An investigation needs independently measured speed and radial force, with mass and radius controlled. Use an appropriately secured apparatus at a controlled rate, a calibrated force instrument with a suitable range, and a clear account of the forces it measures.
A measured total tension is not automatically the radial resultant, as the conical example shows. A hanging mass's weight is not automatically the force transmitted to the moving body either: establish the apparatus force balance and account for intervening friction before equating them.
A fitted intercept can prompt a check for force-zero or background-force errors; scatter alone does not identify its cause. A changing radius or rotation rate can also spoil the intended comparison. Repeats help assess consistency but cannot repair the wrong force model.