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Question W: Derive the dimension of the volume of a cylinder using height (h), radius (r), and diameter (d).

Mathematics

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i

2.14 m/s2.14 \, \text{m/s}

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Question W: Derive the dimension of the volume of a cylinder using height (h), radius (r), and diameter (d).

Step 1: Recall the formula for the volume of a cylinder.
The volume of a cylinder (VV) is given by:

V=πr2hV = \pi r^2 h

Step 2: Identify the dimensions of each physical quantity.

  • Radius (rr) is a length, so its dimension is [L][L].
  • Height (hh) is a length, so its dimension is [L][L].
  • Diameter (dd) is also a length, so its dimension is [L][L].
  • π\pi is a dimensionless constant.

Step 3: Substitute the dimensions into the volume formula.
The dimension of VV is:

[V]=[r]2[h][V] = [r]^2 [h] [V]=[L]2[L][V] = [L]^2 [L] [V]=[L3][V] = [L^3]

Step 4: Express the dimension using diameter (d).
Since the radius rr is half the diameter dd (r=d2r = \frac{d}{2}), its dimension is still [L][L]. Therefore, using diameter instead of radius does not change the overall dimension of volume.

[V]=[d]2[h][V] = [d]^2 [h] [V]=[L]2[L][V] = [L]^2 [L] [V]=[L3][V] = [L^3]

The dimension of the volume of a cylinder is [L3]\boxed{[L^3]}.


Part 4

If a bob is attached to a string of length 80 cm and lifted at an angle of 45 degrees to cause a displacement.

Part i

2.14 m/s2.14 \, \text{m/s}

Find the speed of the moving bob.

Assumption: We assume the question asks for the speed of the bob at the lowest point of its swing after being released from rest at an angle of 45∘45^\circ. We will use the principle of conservation of mechanical energy.

  1. Convert the string length to meters.

    Length of the string, L=80 cm=0.80 mL = 80 \, \text{cm} = 0.80 \, \text{m}.

  2. Calculate the vertical height (hh) the bob falls.

    When the bob is lifted to an angle θ\theta, its vertical height from the pivot is Lcos⁡θL \cos\theta. When it is at the lowest point, its vertical height from the pivot is LL.
    The change in vertical height (the height it falls) is:

    h=L−Lcos⁡θ=L(1−cos⁡θ)h = L - L \cos\theta = L(1 - \cos\theta)

    Given θ=45∘\theta = 45^\circ:

    h=0.80 m(1−cos⁡45∘)h = 0.80 \, \text{m} (1 - \cos 45^\circ) h=0.80 m(1−22)h = 0.80 \, \text{m} \left(1 - \frac{\sqrt{2}}{2}\right) h=0.80 m(1−0.7071)h = 0.80 \, \text{m} (1 - 0.7071) h=0.80 m(0.2929)h = 0.80 \, \text{m} (0.2929) h=0.23432 mh = 0.23432 \, \text{m}
  3. Apply the conservation of mechanical energy.

    At the initial position (lifted at 45∘45^\circ), the bob has potential energy (PEiPE_i) and zero kinetic energy (KEi=0KE_i = 0).
    At the lowest point, the bob has zero potential energy (PEf=0PE_f = 0) and maximum kinetic energy (KEfKE_f).

    PEi+KEi=PEf+KEfPE_i + KE_i = PE_f + KE_f mgh+0=0+12mv2mgh + 0 = 0 + \frac{1}{2}mv^2

    Where mm is the mass of the bob, gg is the acceleration due to gravity (9.8 m/s29.8 \, \text{m/s}^2), and vv is the speed at the lowest point.

  4. Solve for the speed (vv).

    mgh=12mv2mgh = \frac{1}{2}mv^2

    Divide both sides by mm:

    gh=12v2gh = \frac{1}{2}v^2 v2=2ghv^2 = 2gh v=2ghv = \sqrt{2gh}

    Substitute the values for gg and hh:

    v=2×9.8 m/s2×0.23432 mv = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 0.23432 \, \text{m}} v=4.592672 m2/s2v = \sqrt{4.592672 \, \text{m}^2/\text{s}^2} v≈2.143 m/sv \approx 2.143 \, \text{m/s}

    The speed of the moving bob at the lowest point is 2.14 m/s\boxed{2.14 \, \text{m/s}}.

Part ii

Calculate X from the diagram shown.

The diagram mentioned in this part of the question is not provided. Therefore, it is not possible to calculate X.

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