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Worksheet: Center of Mass of a Composite Lamina in 2D

Q1:

The line 𝑦 = 1 0 is the axis of symmetry of a uniform triangular lamina that is isosceles. The π‘₯ -coordinate of the lamina’s centre of mass is βˆ’ 6 . Given that one of the triangle’s vertices lies at the point ( 3 , 5 ) , determine the coordinates of the other two vertices.

  • A ( 3 , 3 5 ) , ( βˆ’ 2 4 , 1 0 )
  • B ( 3 , 1 5 ) , ( βˆ’ 1 2 , 1 0 )
  • C ( 3 , 1 0 ) , ( βˆ’ 1 2 , 1 0 )
  • D ( 3 , 1 5 ) , ( βˆ’ 2 4 , 1 0 )

Q2:

Find the angle that 𝐴 𝐡 makes with the vertical if this L-shaped uniform lamina is freely suspended from 𝐴 , giving your answer to the nearest degree.

Q3:

Find, to the nearest degree, the angle that the line π‘ˆ 𝑇 makes with the vertical if the given uniform lamina is hung freely from 𝑄 .

Q4:

If the given uniform lamina is freely suspended from the point 𝑂 and hangs in equilibrium, find the angle between 𝑂 𝐴 and the downward vertical correct to one decimal place.

Q5:

The shape in the given figure has been formed by bending a uniform piece of wire. 𝑃 𝑄 𝑆 𝑇 is a rectangle where 𝑇 𝑆 is 12 cm and 𝑄 𝑅 𝑆 is a semicircle with centre 𝑂 and a radius of 13 cm. Find the distance between 𝑂 and the centre of mass of the shape giving your answer accurate to two decimal places.

Q6:

A uniform wire of length 404 cm is shaped into a trapezium 𝐴 𝐡 𝐢 𝐷 such that 𝐴 𝐡 = 1 1 8 c m , 𝐢 𝐷 = 9 0 c m , 𝐴 𝐷 = 9 6 c m , and π‘š ∠ 𝐷 𝐴 𝐡 = π‘š ∠ 𝐢 𝐷 𝐴 = 9 0 ∘ . The wire was suspended from 𝐴 . Find the angle that 𝐴 𝐷 makes with the vertical when it is hanging in its equilibrium position. Round your answer to the nearest minute if necessary.

  • A 5 1 4 9 β€² ∘
  • B 4 6 5 5 β€² ∘
  • C 4 0 8 β€² ∘
  • D 4 9 5 2 β€² ∘

Q7:

The given figure shows an L-shaped framework formed by bending a uniform wire. Find the distance between the framework’s centre of mass and 𝐴 𝐡 and the distance between the centre of mass and 𝐴 𝐹 .

  • A 5.4 cm from 𝐴 𝐡 , 2.34 cm from 𝐴 𝐹
  • B 2.14 cm from 𝐴 𝐡 , 2.34 cm from 𝐴 𝐹
  • C 2.34 cm from 𝐴 𝐡 , 5.9 cm from 𝐴 𝐹
  • D 5.4 cm from 𝐴 𝐡 , 5.9 cm from 𝐴 𝐹
  • E 2.14 cm from 𝐴 𝐡 , 5.9 cm from 𝐴 𝐹

Q8:

A uniform lamina is shaped as a parallelogram 𝐴 𝐡 𝐢 𝐷 such that 𝐴 𝐡 = 1 7 4 c m , 𝐡 𝐢 = 8 7 c m , and ∠ 𝐡 𝐢 𝐴 = 9 0 ∘ . The lamina is suspended from the point 𝐸 on 𝐢 𝐷 which causes 𝐢 𝐷 to be horizontal when the system is hanging in equilibrium. Find the length of 𝐢 𝐸 .

  • A 261 cm
  • B 8 7 4 cm
  • C 2 6 1 2 cm
  • D 2 6 1 4 cm

Q9:

A uniform square lamina 𝐴 𝐡 𝐹 𝐺 is joined to a smaller uniform square lamina 𝐡 𝐢 𝐷 𝐸 whose density is 2 4 7 times that of the first. The composite body is freely suspended from 𝐺 . Determine the tangent of the angle that  𝐺 𝐴 makes with the vertical when the body is hanging in its equilibrium position.

  • A 1 1 4
  • B 8 1 1
  • C 1 6 1 1
  • D 1 1 8

Q10:

A uniform wire is bent into an isosceles trapezium 𝐴 𝐡 𝐢 𝐷 , where 𝐴 𝐷 β«½ 𝐡 𝐢 , 𝐴 𝐷 = 1 1 c m , 𝐢 𝐷 = 1 7 c m , and 𝐡 𝐢 = 2 7 c m . The trapezium is located in the first quadrant of a Cartesian plane such that point 𝐢 is at the origin and point 𝐡 is on the π‘₯ -axis. Determine the coordinates of the centre of gravity of the wire. The wire was then suspended freely from 𝐴 . Find the measure of the angle of inclination πœƒ of 𝐴 𝐷 to the vertical when the body is hanging in its equilibrium position rounding your answer to the nearest minute.

  • A ο€Ό 1 3 5 1 6 , 1 9 9 7 2  , πœƒ = 1 8 8 β€² ∘
  • B ο€Ό 2 7 2 , 3 5 6  , πœƒ = 2 3 2 2 β€² ∘
  • C ο€Ό 1 3 5 1 6 , 1 9 9 7 2  , πœƒ = 7 1 5 2 β€² ∘
  • D ο€Ό 2 7 2 , 3 5 6  , πœƒ = 5 9 2 β€² ∘

Q11:

A thin steel frame is in the form of a trapezium in which , , and . The frame is located in a Cartesian plane such that is at the origin and is on the -axis. The section is made of a metal whose density is twice that of the metal used in the remaining part of the frame. Determine coordinates of the centre of gravity of the frame.

  • A
  • B
  • C
  • D

Q12:

The figure shows a uniform square lamina 𝐴 𝐡 𝐢 𝐷 of side length 76 cm and mass π‘š . The centre of the square is 𝑀 , and 𝐸 and 𝐹 are the midpoints of 𝐴 𝐷 and 𝐷 𝐢 respectively. The corner 𝐸 𝐹 𝐷 was bent over so that 𝐷 meets 𝑀 . Find the tangent of the angle πœƒ that  𝐴 𝐢 makes to the vertical when the body is hanging in its equilibrium position from 𝐴 . A weight of mass π‘š 2 was later added to the lamina, causing the centre of mass of the system to be at the centre of the square. Find the coordinates of the point where the weight was added.

  • A t a n πœƒ = 2 3 2 5 , ο€Ό 4 3 7 1 2 , 4 3 7 1 2 
  • B t a n πœƒ = 2 5 2 3 , ο€Ό 2 4 7 6 , 2 4 7 6 
  • C t a n πœƒ = 2 5 2 3 , ο€Ό 4 3 7 1 2 , 4 3 7 1 2 
  • D t a n πœƒ = 2 3 2 5 , ο€Ό 2 4 7 6 , 2 4 7 6 