Question 11 of 18intermediate🔧 ApplyNumerical3 marks

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

Correct Answer

Given: Radius R = 80 cm = 0.8 m, Number of revolutions = 14, Time = 25 s

Step 1: Calculate frequency Frequency (ν) = Number of revolutions / Time = 14/25 = 0.56 Hz

Step 2: Calculate angular speed Angular speed (ω) = 2πν = 2π × 0.56 = 3.52 rad/s

Step 3: Calculate centripetal acceleration a_c = ω²R = (3.52)² × 0.8 = 9.9 m/s²

Direction: The acceleration is always directed towards the centre of the circular path (centripetal acceleration). Since the direction changes continuously, acceleration is not a constant vector, but its magnitude remains constant at 9.9 m/s².

Exercise: EXERCISES | Q: 3.14 | (Page 21)
For More Understanding

Explanation

This question tests understanding of uniform circular motion. The stone undergoes centripetal acceleration directed towards the centre. Using the formula a_c = ω²R where ω = 2πν and ν = frequency, we calculate the magnitude. The context from Example 3.9 shows similar calculations for uniform circular motion problems.

Solution Steps

  1. Step 1: Calculate frequency ν = 14/25 = 0.56 Hz

  2. Step 2: Calculate angular speed ω = 2πν = 3.52 rad/s

  3. Step 3: Calculate centripetal acceleration a_c = ω²R = 9.9 m/s²

  4. Step 4: State direction - towards the centre of the circle

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