a uniform rod of length 1 metre is bent at its midpoint to make 90 angle. distance of centre of mass from centre of rod is
Answers
Answer:
Explanation:
=> Here, a uniform rod of length 1 metre is bent at its midpoint to make 90 angle.
Thus, length of rod, L = 1 metre
mass of rod, (m1 + m2) = 2m
when rod is bent, the length will be half of rod length (L/2).
∴ coordinates on x axis - (x1, y1) = (L/2, 0)
coordinates on y axis - (x2, y2) = (0, L/2)
=> Centre of mass from centre of rod is:
x_centre of mass = m1x1 + m2x2 / m1 + m2
= m*(L/2) + m*(0) / 2m
= L/4
y_centre of mass = m1y1 + m2y2 / m1 + m2
= m*(0) + m*(L/2) / 2m
= L/4
Thus, Centre of mass (x, y) = (L/4, L/4)
=> Now, distance of centre of mass from centre of rod is:
d = √x² + y²
= √ (L/4)² + (L/4)²
= √ (L²/16) + (L²/16)
= √2(L²/16)
= L/4 * √2
= 1 / 2√2
= 0.353 m
Thus, the distance of centre of mass from centre of rod is 0.353 m.
Answer:a uniform rod of length 1 metre is bent at its midpoint to make 90 angle.
Thus, length of rod, L = 1 metre
mass of rod, (m1 + m2) = 2m
when rod is bent, the length will be half of rod length (L/2).
∴ coordinates on x axis - (x1, y1) = (L/2, 0)
coordinates on y axis - (x2, y2) = (0, L/2)
=> Centre of mass from centre of rod is:
x_centre of mass = m1x1 + m2x2 / m1 + m2
= m×(L/2) + m×(0) / 2m
= L/4
y_centre of mass = m1y1 + m2y2 / m1 + m2
= m*(0) + m×(L/2) / 2m
= L/4
Thus, Centre of mass (x, y) = (L/4, L/4)
=> Now, distance of centre of mass from centre of rod is:
d = √x² + y²
= √ (L/4)² + (L/4)²
= √ (L²/16) + (L²/16)
= √2(L²/16)
= L/4 × √2
= 1 / 2√2
= 0.353 m
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