Circular Motion: Centripetal Acceleration & Tension

Physics Motion 11th-12th Grade
PROBLEM
An athlete whirls a 6.51 kg hammer tied to the end of a 1.3 m chain in a simple horizontal circle. The hammer moves at the rate of 1.77 rev/s. What is the centripetal acceleration of the hammer? Answer in units of m/s². What is the tension in the chain? Answer in units of N. (Assume the athlete's arm length is included in the chain length.)

What This Problem Teaches

  • Converting between rotational frequency (rev/s) and linear velocity in circular motion
  • Applying the centripetal acceleration formula: ac = v²/r
  • Understanding the relationship between centripetal force and tension in horizontal circular motion
  • Working with multiple connected physics concepts in a single problem
  • Developing intuition for the magnitudes involved in rotational motion

Visualizing the Setup

An athlete whirls a 6.51 kg hammer tied to the end of a 1.3 m chain in a simple horizontal circle. The hammer moves...

The hammer follows a circular path of radius 1.3 m. The velocity is always tangent to the circle, while the centripetal acceleration (and tension force) points toward the center.

Solution: The Linear Velocity Approach

Step 1 — Convert frequency to linear speed

The hammer completes 1.77 revolutions per second. In one revolution, it travels a distance equal to the circumference of the circle:

Circumference = 2πr = 2π(1.3 m) = 8.168 m

The linear speed is frequency times circumference:

v = f × 2πr = 1.77 rev/s × 8.168 m/rev = 14.46 m/s

Step 2 — Calculate centripetal acceleration

Centripetal acceleration is given by the formula ac = v²/r:

ac = v²/r = (14.46 m/s)² / 1.3 m = 209.1 m²/s² / 1.3 m = 160.8 m/s²

Step 3 — Find the tension in the chain

In horizontal circular motion, the chain tension provides the centripetal force. Using Newton's second law:

Fc = mac = T T = (6.51 kg)(160.8 m/s²) = 1047 N

Solution: Method 2 — The Angular Velocity Route

Step 1 — Convert frequency to angular velocity

Angular velocity ω is related to frequency by ω = 2πf:

ω = 2πf = 2π(1.77 rad/s) = 11.12 rad/s

Step 2 — Use the rotational form of centripetal acceleration

Centripetal acceleration can also be expressed as ac = ω²r:

ac = ω²r = (11.12 rad/s)² × 1.3 m = 123.7 × 1.3 = 160.8 m/s²

Step 3 — Calculate tension

The tension calculation remains the same:

T = mac = (6.51 kg)(160.8 m/s²) = 1047 N
Final Answers:
Centripetal acceleration: 160.8 m/s²
Chain tension: 1047 N

Verification

Let's verify our answers by checking the dimensional analysis and magnitude:

Dimensional Check

ac = v²/r = (m/s)²/(m) = m²/s²/m = m/s² ✓ T = ma = kg × m/s² = kg⋅m/s² = N ✓

Magnitude Check

The centripetal acceleration of 160.8 m/s² is about 16 times gravitational acceleration (16g). This is reasonable for an athletic hammer throw at high rotational speed.

The tension of 1047 N corresponds to supporting a weight of about 107 kg (235 lbs), which is plausible for a strong athlete whirling a heavy hammer at high speed.

Alternative Verification

We can verify by using the alternate formula ac = 4π²f²r:

ac = 4π²f²r = 4π²(1.77)²(1.3) = 39.48 × 3.13 × 1.3 = 160.8 m/s² ✓

Watch Out For These Mistakes

✗ Using frequency directly in ac = v²/r
Some students write ac = (1.77)²/1.3 = 2.4 m/s². This treats frequency as if it were velocity, ignoring the need to convert rev/s to m/s.
✗ Forgetting to square the velocity
Writing ac = v/r = 14.46/1.3 = 11.1 m/s² misses the quadratic relationship. Centripetal acceleration increases with the square of speed, not linearly.
✗ Including gravity in horizontal circular motion
Some students write T = mac + mg, but in horizontal circular motion, gravity acts vertically while tension acts horizontally. These forces are perpendicular, not additive.
✗ Unit confusion with angular velocity
Using ω = 1.77 rad/s directly instead of ω = 2π(1.77) = 11.12 rad/s. Remember: 1 revolution = 2π radians.

The Physics Behind This

This problem demonstrates several fundamental principles working together:

The Centripetal Acceleration Formula: ac = v²/r = ω²r = 4π²f²r

These three forms are equivalent and useful in different contexts. The v²/r form is intuitive when you know linear speed. The ω²r form is natural when working with angular motion. The 4π²f²r form is convenient when frequency is given directly.

The key insight is that circular motion requires constant acceleration toward the center, even though the speed is constant. This acceleration must be provided by some force—in this case, the tension in the chain.

The magnitude of this acceleration depends quadratically on the rotation rate. Double the frequency, and you quadruple both the centripetal acceleration and the required tension force.

Real Applications

  • Engineering: Designing centrifuges for medical labs, where the centripetal acceleration can reach thousands of g's to separate blood components
  • Automotive: Calculating maximum safe cornering speeds for vehicles, where tire friction provides the centripetal force
  • Aerospace: Determining structural loads on rotating spacecraft components or helicopter rotors
  • Athletics: Optimizing technique in hammer throw, discus, and other rotational sports

Four "What-If?" Problems

1
Different Frequency
An athlete whirls the same 6.51 kg hammer on a 1.3 m chain, but at 2.2 rev/s instead of 1.77 rev/s. What is the new centripetal acceleration and chain tension?
Step 1 — Find the linear speed

v = 2πrf = 2π(1.3)(2.2) = 18.0 m/s

Step 2 — Calculate centripetal acceleration

ac = v²/r = (18.0)²/1.3 = 324/1.3 = 249.2 m/s²

Step 3 — Find tension

T = mac = (6.51)(249.2) = 1622 N

Verification

The frequency increased by factor 2.2/1.77 = 1.24, so acceleration should increase by (1.24)² = 1.55. Indeed, 249.2/160.8 ≈ 1.55 ✓

Answer: ac = 249.2 m/s², T = 1622 N

2
Vertical Circle Challenge
The athlete whirls the hammer in a vertical circle at the same 1.77 rev/s. What is the tension in the chain at the bottom of the swing? (Hint: Now gravity matters!)
Step 1 — Centripetal acceleration is the same

ac = 160.8 m/s² (same calculation as before)

Step 2 — Analyze forces at bottom

At the bottom, both tension and weight point toward the center (upward): T - mg = mac

Step 3 — Calculate tension

T = mac + mg = (6.51)(160.8) + (6.51)(9.8) = 1047 + 63.8 = 1111 N

Verification

The tension is higher than horizontal motion by exactly the weight (mg = 63.8 N), which makes physical sense.

Answer: T = 1111 N

3
Maximum Safe Frequency
If the chain can withstand a maximum tension of 1500 N before breaking, what is the maximum safe frequency for whirling the 6.51 kg hammer horizontally with radius 1.3 m?
Step 1 — Find maximum centripetal acceleration

ac,max = Tmax/m = 1500/6.51 = 230.4 m/s²

Step 2 — Find maximum linear speed

vmax = √(ac,max × r) = √(230.4 × 1.3) = √299.5 = 17.31 m/s

Step 3 — Convert to frequency

fmax = vmax/(2πr) = 17.31/(2π × 1.3) = 17.31/8.17 = 2.12 rev/s

Verification

Check: ac = 4π²f²r = 4π²(2.12)²(1.3) = 230.4 m/s², so T = 6.51 × 230.4 = 1500 N ✓

Answer: fmax = 2.12 rev/s

4
Working Backward from Measurements
An athlete whirls a 7.2 kg hammer. The measured centripetal acceleration is 180 m/s² and the chain tension is 1296 N. What is the radius of the circular path, and at what frequency is the hammer rotating?
Step 1 — Verify consistency

Check if T = mac: 1296 = 7.2 × 180 = 1296 N ✓

Step 2 — Find radius from ac = v²/r

We need another relationship. Use ac = ω²r, but first find ω from ac = 4π²f²r

Step 3 — Use v²/r = ω²r to find v, then r

From ac = v²/r, if we assume r = 1.5 m: v = √(180 × 1.5) = √270 = 16.43 m/s
Check: f = v/(2πr) = 16.43/(2π × 1.5) = 1.74 rev/s

Verification

ac = 4π²f²r = 4π²(1.74)²(1.5) = 180 m/s² ✓

Answer: r = 1.5 m, f = 1.74 rev/s

Frequently Asked Questions

Centripetal acceleration is the inward acceleration that keeps an object moving in a circular path. It's calculated as ac = v²/r where v is the tangential speed and r is the radius. In this problem, the hammer moving at 14.46 m/s with radius 1.3 m experiences ac = 160.8 m/s². This acceleration doesn't change the speed—it only changes the direction to keep the object in a circle.
Multiply the frequency by the circumference of the circle: v = f × 2πr. Here, 1.77 rev/s × 2π × 1.3 m = 14.46 m/s. Each revolution covers a distance of 2πr, so the frequency tells you how many of those distances are covered per second. This is why the circumference appears in the conversion.
In horizontal circular motion, gravity acts vertically downward while the chain tension provides the horizontal centripetal force. Since these forces are perpendicular, the tension must equal mac to maintain the circular path. In this problem, T = 6.51 kg × 160.8 m/s² = 1047 N. The athlete's arms must provide this force to keep the hammer moving in a circle.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-08-17

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