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Stopper on a string: period to speed to the tension

SPH4U Grade 12 Physics · Dynamics

For the classic circular-motion lab, a student whirls a 250 g rubber stopper on a cord in a horizontal circle of radius 85.0 cm above their head. A partner times 20 complete revolutions at 8.40 s, so each lap takes 0.420 s. Treat the circle as horizontal and the cord as massless.

Given
  • m = 250 g — Mass of the stopper
  • r = 85 cm — Radius of the circle
  • T = 0.42 s — Period (20 laps in 8.40 s)
Determine
  1. (a)the speed of the stopper along its circle
  2. (b)its centripetal acceleration
  3. (c)the tension in the cord
  4. (d)the stopper's weight, for comparison with that tension
Step 1 of 4(a) · solve for Speed

One circumference per period: v = 2πr/T. The two unit fixes both happen here — the 85.0 cm radius becomes 0.850 m, and the period is 8.40 s divided by the 20 laps, never the full 8.40 s, which would slow the stopper twenty-fold.

Rearranged for v
v=2πrTv = \frac{2\pi r}{T}
Your values, in your units
v=2(3.14159) (85 cm)(0.42 s)v = \frac{2\left( 3.14159 \right) \, \left( 85\ \text{cm} \right)}{\left( 0.42\ \text{s} \right)}
Converted to base units
v=2(3.14159) (0.85 m)(0.42 s)v = \frac{2\left( 3.14159 \right) \, \left( 0.85\ \text{m} \right)}{\left( 0.42\ \text{s} \right)}
Answer
v=12.716 m/sv = 12.716\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Speed in Circular Motion (v = 2πr/T) solver →

Step 2 of 4(b) · solve for Centripetal acceleration

"Constant speed" is not "no acceleration": the velocity's direction is changing every instant, and a = v²/r measures how hard. At 190 m/s² this stopper is being yanked toward the centre at more than 19 g — the square on the v is what makes whirling things violent.

Rearranged for ac
ac=v2ra_c = \frac{v^2}{r}
12.716 m/scarried from step 1
Your values, in your units
ac=(12.716 m/s)2(85 cm)a_c = \frac{\left( 12.716\ \text{m/s} \right)^2}{\left( 85\ \text{cm} \right)}
Converted to base units
ac=(12.716 m/s)2(0.85 m)a_c = \frac{\left( 12.716\ \text{m/s} \right)^2}{\left( 0.85\ \text{m} \right)}
Answer
ac=190.23 m/s2a_c = 190.23\ \text{m/s}^{2}

Carried onward at full precision, not this rounded figure.

Open the Centripetal Acceleration (a = v²/r) solver →

Step 3 of 4(c) · solve for Centripetal force

Centripetal force is not a new force on the diagram — it is the job description the cord's tension is filling. F = mv²/r with the 250 g in kilograms names the pull the cord must supply, and if the cord snaps the stopper leaves along the tangent, not outward.

Rearranged for Fc
Fc=mv2rF_c = \frac{m v^2}{r}
12.716 m/scarried from step 1
Your values, in your units
Fc=(250 g) (12.716 m/s)2(85 cm)F_c = \frac{\left( 250\ \text{g} \right) \, \left( 12.716\ \text{m/s} \right)^2}{\left( 85\ \text{cm} \right)}
Converted to base units
Fc=(0.25 kg) (12.716 m/s)2(0.85 m)F_c = \frac{\left( 0.25\ \text{kg} \right) \, \left( 12.716\ \text{m/s} \right)^2}{\left( 0.85\ \text{m} \right)}
Answer
Fc=47.558 NF_c = 47.558\ \text{N}

Carried onward at full precision, not this rounded figure.

Open the Centripetal Force (F = mv²/r) solver →

Step 4 of 4(d) · solve for Weight

The same stopper hanging still would load the cord with just mg. Set 2.45 N beside step 3's 47.6 N: the whirl multiplies the cord's burden about nineteen-fold, which is why the lab manual says check the knot.

Rearranged for W
W=mgW = m g
Your values, in your units
W=(250 g) (9.80665 m/s2)W = \left( 250\ \text{g} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)
Converted to base units
W=(0.25 kg) (9.80665 m/s2)W = \left( 0.25\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)
Answer
W=2.4517 NW = 2.4517\ \text{N}

Carried onward at full precision, not this rounded figure.

Open the Weight (W = mg) solver →

Therefore the stopper circles at 12.7 m/s, accelerates toward the centre at 190 m/s², and loads the cord with 47.6 N — about nineteen times its own 2.45 N weight.

Why this order

The chain runs geometry → kinematics → dynamics, and each arrow is one formula. The period is the only thing actually measured, so everything begins there; v = 2πr/T is pure geometry (a distance per lap over a time per lap); a = v²/r is kinematics, true of any object on that circle regardless of what it is; and only in step 3 does the stopper's mass — the dynamics — enter at all. The error this order is designed to expose is the phrase "centripetal force" treated as a force of its own. Nothing new pulls on the stopper: the cord's tension IS the centripetal force here, the way friction is for a cornering car and gravity is for the Moon. Adding an extra outward "centrifugal force" to balance it produces a stopper in equilibrium, which would oblige it to travel in a straight line — precisely what it is not doing.

The factor of nineteen between tension and weight deserves the pause. It comes almost entirely from the v² in the numerator: halve the period and the tension quadruples, which a hand on the cord can feel immediately and which is why this lab is usually done with a glass tube and a safety washer. The comparison in part (d) also quietly justifies the "horizontal circle" idealization — the cord must in fact sag a few degrees so a component of tension can carry the 2.45 N of weight, but with the tension nineteen times larger, that angle is about 3° and the horizontal treatment is honest to about one part in seven hundred. Checking the size of what you neglected, after you neglected it, is the mark of a defensible approximation.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.