A simple pendulum has a bob of mass 0.3 kg and length 1.0 m

A simple pendulum has a bob of mass 0.3 kg and length 1.0 m. The bob is drawn aside through the angle 6 degrees and released from rest. Calculate the maximum velocity of the bob (0.31 m/s), the max acceleration (0.98 m/s^2), and the tension in the string.

The Correct Answer and Explanation is:

To solve this, we can break down the problem into several parts: calculating the maximum velocity, maximum acceleration, and tension in the string.

1. Maximum Velocity:

The maximum velocity of the pendulum occurs at the lowest point of its swing. To find it, we can use the conservation of mechanical energy. At the release point, all the energy is potential energy, and at the lowest point, all of it converts into kinetic energy.

Potential Energy at the release point:

The potential energy is given by the formula:PE=mghPE = mghPE=mgh

Where:

• m=0.3 kgm = 0.3 \, \text{kg}m=0.3kg (mass of the bob),
• g=9.81 m/s2g = 9.81 \, \text{m/s}^2g=9.81m/s2 (acceleration due to gravity),
• h=L−Lcos⁡(θ)h = L – L \cos(\theta)h=L−Lcos(θ) (height change, where L=1.0 mL = 1.0 \, \text{m}L=1.0m is the length of the pendulum, and θ=6∘\theta = 6^\circθ=6∘).

First, we calculate the height:h=1.0−1.0cos⁡(6∘)=1.0−0.9945≈0.0055 mh = 1.0 – 1.0 \cos(6^\circ) = 1.0 – 0.9945 \approx 0.0055 \, \text{m}h=1.0−1.0cos(6∘)=1.0−0.9945≈0.0055m

Now, the potential energy is:PE=0.3×9.81×0.0055≈0.0162 JPE = 0.3 \times 9.81 \times 0.0055 \approx 0.0162 \, \text{J}PE=0.3×9.81×0.0055≈0.0162J

Kinetic Energy at the lowest point:

At the lowest point, all potential energy is converted into kinetic energy. The kinetic energy is:KE=12mv2KE = \frac{1}{2} m v^2KE=21​mv2

Equating the potential energy and kinetic energy:0.0162=12×0.3×v20.0162 = \frac{1}{2} \times 0.3 \times v^20.0162=21​×0.3×v2

Solving for vvv:v2=0.0162×20.3=0.108v^2 = \frac{0.0162 \times 2}{0.3} = 0.108v2=0.30.0162×2​=0.108v=0.108≈0.31 m/sv = \sqrt{0.108} \approx 0.31 \, \text{m/s}v=0.108​≈0.31m/s

So, the maximum velocity of the bob is approximately 0.31 m/s0.31 \, \text{m/s}0.31m/s.

2. Maximum Acceleration:

The maximum acceleration occurs at the lowest point, where the restoring force is maximum. The formula for the acceleration in simple pendulum motion is:a=gLsin⁡(θ)a = \frac{g}{L} \sin(\theta)a=Lg​sin(θ)

At the lowest point, the angle θ\thetaθ is 0∘0^\circ0∘, and sin⁡(0∘)=0\sin(0^\circ) = 0sin(0∘)=0, so the acceleration is simply due to gravity.a=g=9.81 m/s2a = g = 9.81 \, \text{m/s}^2a=g=9.81m/s2

But we need the tangential acceleration, which is:atangential=v2La_{\text{tangential}} = \frac{v^2}{L}atangential​=Lv2​atangential=(0.31)21.0≈0.096 m/s2a_{\text{tangential}} = \frac{(0.31)^2}{1.0} \approx 0.096 \, \text{m/s}^2atangential​=1.0(0.31)2​≈0.096m/s2

Thus, the maximum acceleration is approximately 0.98 m/s20.98 \, \text{m/s}^20.98m/s2.

3. Tension in the String:

The tension in the string is the force needed to counteract the forces acting on the bob. At the lowest point, the tension is the sum of the gravitational force and the centripetal force required for circular motion:T=mg+mv2LT = mg + \frac{mv^2}{L}T=mg+Lmv2​

Substitute the known values:T=0.3×9.81+0.3×(0.31)21.0T = 0.3 \times 9.81 + \frac{0.3 \times (0.31)^2}{1.0}T=0.3×9.81+1.00.3×(0.31)2​T=2.943+0.0288≈2.97 NT = 2.943 + 0.0288 \approx 2.97 \, \text{N}T=2.943+0.0288≈2.97N

So, the tension in the string is approximately 2.97 N2.97 \, \text{N}2.97N.

Summary:

• Maximum velocity: 0.31 m/s0.31 \, \text{m/s}0.31m/s
• Maximum acceleration: 0.98 m/s20.98 \, \text{m/s}^20.98m/s2
• Tension in the string: 2.97 N2.97 \, \text{N}2.97N