Figure shows a wheel rotating about Q2- Two points A and B located along the radius of wheel have speeds of 80ms and 140ms respectively- The distance between the points A and B is 300 mm- The diameter of the wheel in mm is

Angular velocity of point A and B will be same Where, R = radius of the wheel ω_A=ω_B V_AR 0.3=V_BR 80R 0.3=140R 80 R = 140 R 42 42 = 60 R R = 0.7 m ∴ Diameter ...

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Standard XII

Physics

Introduction to Radial & Tangential Acceleration

Figure shows ...

Question

Figure shows a wheel rotating about $Q_{2} .$ Two points A and B located along the radius of wheel have speeds of $80 \frac{m}{s}$ and $140 \frac{m}{s}$ respectively. The distance between the points A and B is 300 mm. The diameter of the wheel (in mm) is

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Solution

Angular velocity of point A and B will be same

Where, R = radius of the wheel $ω_{A} = ω_{B}$
$\frac{V_{A}}{R - 0.3} = \frac{V_{B}}{R}$
$\frac{80}{R - 0.3} = \frac{140}{R}$
$80 R = 140 R - 42$
$42 = 60 R$
$R = 0.7 m$
$∴$ Diameter of the wheel,
$D = 2 R = 1.4 m = 1400 m m$

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Figure shows a wheel rotating about Q2. Two points A and B located along the radius of wheel have speeds of 80ms and 140ms respectively. The distance between the points A and B is 300 mm. The diameter of the wheel (in mm) is

Angular velocity of point A and B will be same Where, R = radius of the wheel ωA=ωB VAR−0.3=VBR 80R−0.3=140R 80R=140R−42 42=60R R=0.7m ∴ Diameter of the wheel, D=2R=1.4m=1400mm

Figure shows a wheel rotating about $Q_{2} .$ Two points A and B located along the radius of wheel have speeds of $80 \frac{m}{s}$ and $140 \frac{m}{s}$ respectively. The distance between the points A and B is 300 mm. The diameter of the wheel (in mm) is

Angular velocity of point A and B will be same

Where, R = radius of the wheel $ω_{A} = ω_{B}$
$\frac{V_{A}}{R - 0.3} = \frac{V_{B}}{R}$
$\frac{80}{R - 0.3} = \frac{140}{R}$
$80 R = 140 R - 42$
$42 = 60 R$
$R = 0.7 m$
$∴$ Diameter of the wheel,
$D = 2 R = 1.4 m = 1400 m m$