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Worksheet: Center of Mass of Laminas

Q1:

𝐴 𝐡 is a nonuniform rod of length 108 cm and mass 675 g. Two weights of 50 g and 350 g are attached to the ends 𝐴 and 𝐡 , respectively, causing the centre of gravity of the system to be at the midpoint of the rod. Determine the distance between the end 𝐴 and the centre of gravity of the rod.

Q2:

The figure shows a uniform wire 𝐴 𝐡 𝐢 𝐷 of length 10 cm where 𝐴 𝐡 = 𝐡 𝐢 = 2 𝐢 𝐷 = 4 c m . Find the coordinates of the center of gravity of the wire with respect to axes  𝐡 𝐴 and οƒͺ 𝐡 𝐢 .

  • A ο€Ό 1 6 5 , 2 
  • B ο€Ό 3 5 , 8 5 
  • C ο€Ό 2 , 1 6 5 
  • D ο€Ό 8 5 , 3 5 

Q3:

A uniform lamina 𝐴 𝐡 𝐢 𝐷 is shaped as a trapezoid, where π‘š ∠ 𝐴 = π‘š ∠ 𝐷 = 9 0 ∘ , 𝐢 𝐷 = 1 0 c m , 𝐴 𝐷 = 8 c m , and 𝐴 𝐡 = 2 5 c m . Find the coordinates of the center of gravity of the lamina.

  • A ο€Ό 1 4 5 1 4 , 4 
  • B ο€Ό 3 5 4 , 6 
  • C ο€Ό 1 4 5 8 , 7 
  • D ο€Ό 6 5 7 , 2 4 7 

Q4:

A uniform rectangular board had an area of 385 cm2. A circular hole of area 66 cm2 was cut out of the board such that the distance between the center of the board 𝐸 and the center of the circular hole 𝑂 was 6 cm. Find the distance between the center of gravity of the system and the center of the board.

  • A 6 4 1 cm
  • B 3 6 4 1 cm
  • C 6 2 9 cm
  • D 3 6 2 9 cm

Q5:

A blacksmith is crafting the shown shape by removing three semicircles of radius 9 cm from a larger semicircle of radius 27 cm. Find the coordinates of the centre of mass of the shape, assuming that it is a uniform lamina.

  • A ο€Ό 0 , 1 6 0 3 πœ‹ 
  • B ο€Ό 0 , 8 0 3 πœ‹ 
  • C ο€Ό 0 , 8 1 πœ‹ 
  • D ο€Ό 0 , 4 8 πœ‹ 
  • E ο€Ό 0 , 1 6 2 πœ‹ 

Q6:

A thin uniform wire is shaped into trapezium 𝐴 𝐡 𝐢 𝐷 where π‘š ∠ 𝐡 = π‘š ∠ 𝐢 = 9 0 ∘ , 𝐴 𝐡 = 4 9 4 c m , 𝐡 𝐢 = 1 0 5 c m , and 𝐢 𝐷 = 1 3 4 c m . Find the distance between the centre of gravity of the wire and point 𝐡 rounding your answer to two decimal places.

Q7:

A uniform wire of length 99 cm is cut in two. A circle of radius 11 cm is formed from the first part, and the second part is bent from its midpoint 𝐡 to form a right triangle 𝐴 𝐡 𝐢 . The two parts are joined together such that 𝐴 𝐡 and 𝐡 𝐢 touch the circle at 𝐾 and 𝐿 respectively. Given that the two parts are coplanar, determine the distance between 𝐴 𝐡 and the center of gravity of the system. Round your answer to one decimal place.

  • A 11.1 cm
  • B 10.8 cm
  • C 4.4 cm
  • D 8.8 cm

Q8:

A uniform lamina is in the form of an isosceles triangle 𝐴 𝐡 𝐢 in which 𝐴 𝐡 = 𝐴 𝐢 . The height of the triangle, represented by the line 𝐴 𝐷 , is 887 cm long. A straight line is draw so that it passes through the center of gravity of the lamina and is parallel to the base 𝐡 𝐢 . It intersects 𝐴 𝐡 and 𝐴 𝐢 at 𝐸 and 𝐹 respectively. Find the distance between point 𝐷 and the center of gravity of the quadrilateral 𝐸 𝐡 𝐢 𝐹 stating your answer to two decimal places.

  • A 157.69 cm
  • B 197.11 cm
  • C 177.4 cm
  • D 137.98 cm

Q9:

A thin uniform wire is bent around the three sides 𝐴 𝐡 , 𝐡 𝐢 , and 𝐢 𝐷 of a square 𝐴 𝐡 𝐢 𝐷 of side length 81 cm. Find the distances 𝑑 𝐴 𝐡 and 𝑑 𝐡 𝐢 between the center of gravity of the wire and the edges 𝐴 𝐡 and 𝐡 𝐢 respectively. The wire was suspended freely from the point 𝐡 . Find the tangent of the angle of inclination of 𝐡 𝐢 to the vertical, t a n πœƒ , when the body is hanging in its equilibrium position.

  • A 𝑑 = 2 7 𝐴 𝐡 c m , 𝑑 = 8 1 2 𝐡 𝐢 c m , t a n πœƒ = 2 3
  • B 𝑑 = 8 1 2 𝐴 𝐡 c m , 𝑑 = 2 7 𝐡 𝐢 c m , t a n πœƒ = 3 2
  • C 𝑑 = 2 7 𝐴 𝐡 c m , 𝑑 = 8 1 2 𝐡 𝐢 c m , t a n πœƒ = 3 2
  • D 𝑑 = 8 1 2 𝐴 𝐡 c m , 𝑑 = 2 7 𝐡 𝐢 c m , t a n πœƒ = 2 3

Q10:

The figure shows a uniform square lamina of side length 16 cm. It is divided into nine congruent squares as shown. Given that the squares 𝐼 , 𝐷 , and 𝐡 have been removed, determine the coordinates of the center of gravity of the remaining parts of the lamina.

  • A ο€Ό 1 5 2 2 1 , 8 
  • B ο€Ό 4 8 7 , 4 8 7 
  • C ο€Ό 8 0 9 , 8 
  • D ( 8 , 8 )

Q11:

A uniform lamina of weight 165 N is in the shape of a trapezoid 𝐴 𝐡 𝐢 𝐷 where π‘š ∠ 𝐡 = π‘š ∠ 𝐢 = 9 0 ∘ , 𝐴 𝐡 = 𝐡 𝐢 = 5 4 c m , and 𝐢 𝐷 = 3 6 c m . The lamina is positioned on a horizontal table such that 𝐢 𝐷 rests on the table top and 𝐴 𝐡 is vertically above 𝐢 𝐷 . Find the coordinates of the center of mass of the lamina and the maximum weight π‘Š that can be suspended from 𝐴 without the lamina tipping over.

  • A ο€Ό 1 8 6 5 , 1 7 1 5  , π‘Š = 1 2 1 N
  • B ο€Ό 1 1 4 5 , 1 4 4 5  , π‘Š = 5 3 9 N
  • C ο€Ό 1 8 6 5 , 1 7 1 5  , π‘Š = 5 3 9 N
  • D ο€Ό 1 1 4 5 , 1 4 4 5  , π‘Š = 1 2 1 N

Q12:

The figure shows a uniform wire . It has been bent at and to form right-angles. The wire was freely suspended from . Find the size of the inclination angle to the vertical when the body is hanging in its equilibrium position. Round your answer to the nearest minute.

  • A
  • B
  • C
  • D

Q13:

A uniform lamina 𝐴 𝐡 𝐢 𝐷 is in the shape of a rhombus which has a side length of 56 cm and π‘š ∠ 𝐡 𝐴 𝐷 = 1 2 0 ∘ . Its two diagonals intersect at 𝑁 . The triangle 𝐴 𝐡 𝑁 was cut out of the lamina. Calculate the distance 𝑑 between 𝑁 and the center of gravity of the resulting lamina. Given that the lamina was then suspended from point 𝐴 , determine the tangent of the angle 𝐴 𝑁 makes to the vertical, t a n πœƒ , when the lamina is hanging in its equilibrium position.

  • A 𝑑 = 2 8 √ 2 9 c m , t a n πœƒ = √ 3
  • B 𝑑 = 5 6 9 c m , t a n πœƒ = √ 3
  • C 𝑑 = 2 8 √ 2 9 c m , t a n πœƒ = 1 1 0
  • D 𝑑 = 5 6 9 c m , t a n πœƒ = √ 3 1 0

Q14:

𝐴 𝐡 𝐢 𝐷 is a rectangular lamina, in which 𝐴 𝐡 = 2 0 c m and 𝐡 𝐢 = 2 1 c m . The weight of the is lamina 30 N, acting at the point of intersection of its diagonals. It is also freely suspended from vertex 𝐷 so that its plane is vertical. If a couple having a moment of 250 Nβ‹…cm starts acting on the lamina perpendicularly to the plane, find the inclination of the diagonal 𝐷 𝐡 to the vertical when the lamina is in equilibrium, giving your answer to the nearest minute.

  • A 5 6 2 7 β€² ∘
  • B 1 6 4 2 β€² ∘
  • C 5 4 5 5 β€² ∘
  • D 3 5 5 β€² ∘

Q15:

A uniform lamina of mass 9 kg is shaped as triangle 𝐴 𝐡 𝐢 where 𝐴 𝐡 = 𝐴 𝐢 and 𝐡 𝐢 equals the height of the triangle which is 14 cm. Masses of 20 kg , 2 kg , 15 kg , and 34 kg are fixed at the points 𝐴 , 𝐡 , 𝐢 , and 𝐷 respectively, where 𝐷 is the midpoint of 𝐴 𝐡 . Find the distance 𝐿 between the center of gravity of the system and point 𝐢 . Given that the system was freely suspended from 𝐢 , find the measure of the angle of inclination πœƒ of 𝐴 𝐢 to the vertical rounding your answer to the nearest minute.

  • A 𝐿 = 3 . 8 3 c m , πœƒ = 1 8 4 9 β€² ∘
  • B 𝐿 = 1 0 . 1 5 c m , πœƒ = 4 3 3 6 β€² ∘
  • C 𝐿 = 3 . 8 3 c m , πœƒ = 4 4 3 7 β€² ∘
  • D 𝐿 = 1 0 . 1 5 c m , πœƒ = 1 9 5 0 β€² ∘

Q16:

A uniform square lamina of weight 30 g is freely suspended from vertex . A weight of 16 g is fixed to vertex . Find the size of the angle of inclination of the square’s diagonal to the vertical when the body is hanging in its equilibrium position. Round the result to the nearest minute.

  • A
  • B
  • C
  • D

Q17:

The figure shows a uniform square lamina of side length 20 cm. It is divided into nine congruent squares as shown. Given that the square 𝐼 has been removed, determine the coordinates of the center of gravity of the resulting the lamina.

  • A ο€Ό 3 2 3 , 3 2 3 
  • B ο€Ό 2 2 3 , 2 2 3 
  • C ο€Ό 1 0 , 5 5 6 
  • D ο€Ό 5 5 6 , 5 5 6 

Q18:

A nonuniform lamina 𝐴 𝐡 𝐢 𝐷 is shaped as a rectangle such that 𝐴 𝐡 = 7 0 c m and 𝐡 𝐢 = 1 4 0 c m . When it is suspended from 𝐷 and settles in its equilibrium position, 𝐷 𝐡 is vertical. When it is suspended from 𝐴 and settles its equilibrium position, 𝐷 𝐡 is horizontal. Determine the distances, 𝑑 𝐴 𝐡 and 𝑑 𝐡 𝐢 , from the center of gravity to the edges 𝐴 𝐡 and 𝐡 𝐢 .

  • A 𝑑 = 2 8 √ 5 𝐴 𝐡 c m , 𝑑 = 1 5 𝐡 𝐢 c m
  • B 𝑑 = 1 4 𝐴 𝐡 c m , 𝑑 = 2 8 𝐡 𝐢 c m
  • C 𝑑 = 1 5 𝐴 𝐡 c m , 𝑑 = 2 8 √ 5 𝐡 𝐢 c m
  • D 𝑑 = 2 8 𝐴 𝐡 c m , 𝑑 = 1 4 𝐡 𝐢 c m

Q19:

The figure shows a uniform circular lamina of radius 5.6 cm and center 𝑀 . A circular disc of radius 2.4 cm and center 𝑁 has been removed from the lamina as shown. Determine the distance in centimeters between 𝑀 and the center of mass of the resulting lamina.

  • A 6 2 5 cm
  • B 9 2 5 cm
  • C 3 6 2 5 cm
  • D 1 8 2 5 cm

Q20:

A non-uniform lamina 𝐴 𝐡 𝐢 is in the form of a triangle which is right-angled at 𝐡 such that π‘š ∠ 𝐴 = 6 0 ∘ and 𝐴 𝐡 = 3 4 c m . When the lamina is freely suspended from 𝐡 , 𝐴 𝐢 is horizontal. When the lamina is suspended freely from 𝐴 , 𝐴 𝐢 is inclined at an angle of 3 0 ∘ to the vertical. Determine the distances, 𝑑 𝐴 𝐡 and 𝑑 𝐡 𝐢 , from the centre of gravity to the edges 𝐴 𝐡 and 𝐡 𝐢 .

  • A 𝑑 = 3 4 √ 3 3 𝐴 𝐡 c m , 𝑑 = 1 7 √ 3 3 𝐡 𝐢 c m
  • B 𝑑 = 1 7 𝐴 𝐡 c m , 𝑑 = 1 7 √ 3 3 𝐡 𝐢 c m
  • C 𝑑 = 1 7 √ 3 3 𝐴 𝐡 c m , 𝑑 = 3 4 √ 3 3 𝐡 𝐢 c m
  • D 𝑑 = 1 7 √ 3 3 𝐴 𝐡 c m , 𝑑 = 1 7 𝐡 𝐢 c m

Q21:

A uniform lamina 𝐴 𝐡 𝐢 is in the form of an equilateral triangle of side length 75 cm. Its center of mass is 𝐺 . The part 𝐴 𝐡 𝐺 𝐢 was removed. Find the coordinates of the center of mass of 𝐺 𝐡 𝐢 . The lamina 𝐺 𝐡 𝐢 was then suspended from 𝐢 . Find the tangent of the angle that 𝐢 𝐡 makes to the vertical, t a n πœƒ , when the lamina is hanging in its equilibrium position.

  • A ο€Ώ 7 5 2 , 2 5 √ 3 2  , t a n πœƒ = 4 √ 3 9
  • B ο€Ώ 7 5 2 , 5 0 √ 3 3  , t a n πœƒ = 3 √ 3 4
  • C ο€Ώ 7 5 2 , 2 5 √ 3 2  , t a n πœƒ = 3 √ 3 4
  • D ο€Ώ 7 5 2 , 5 0 √ 3 3  , t a n πœƒ = 4 √ 3 9

Q22:

A uniform square-shaped lamina 𝐴 𝐡 𝐢 𝐷 of length of 96 cm is joined to another lamina shaped as an isosceles triangle 𝐸 𝐡 𝐢 of a leg length of 80 cm. Determine the distance between the center of gravity of the whole lamina and the center of gravity of the square, stating your answer to two decimal places. The lamina was then suspended freely from 𝐴 . Find the angle πœƒ that 𝐴 𝐷 makes to the vertical when the lamina is hanging in its equilibrium position, giving your answer to the nearest minute.

  • A 𝑑 = 2 2 . 6 7 c m , πœƒ = 5 5 4 9 β€² ∘
  • B 𝑑 = 1 7 . 3 3 c m , πœƒ = 3 6 1 8 β€² ∘
  • C 𝑑 = 2 2 . 6 7 c m , πœƒ = 3 4 1 1 β€² ∘
  • D 𝑑 = 1 7 . 3 3 c m , πœƒ = 5 3 4 2 β€² ∘

Q23:

The mass of a fine lamina of a uniform thickness and density is 160 grams. It is in the form of trapezium ,where , , , and . A uniform rod of mass 20 grams is coincident with completely. Find the coordinates of the centre of mass of the system. If the body is freely suspended from , find the size of the inclination angle that makes with the vertical in the equilibrium position to the nearest minute.

  • A ,
  • B ,
  • C ,
  • D ,
  • E ,

Q24:

A uniform lamina is in the form of a rectangle , where and . Point is located at the intersection of its diagonals and . Triangle is cut out of the lamina and fixed above triangle . Find the size of the angle that makes to the vertical when it is hanging in its equilibrium position from , stating your answer to the nearest minute.

  • A
  • B
  • C
  • D

Q25:

A uniform disk has a radius of 84 cm and a center 𝑀 . A circular hole of radius 63 cm was drilled into the disk such that it touches the outer circle of the disk at a point 𝑋 . Find the distance, 𝑑 , between point 𝑋 and the center of gravity of the resulting disk. The disk was then suspended freely from a point 𝑍 which is where the perpendicular radius to 𝑀 𝑋 meets the edge of the disk. Determine the tangent of the angle that 𝑍 𝑀 makes to the vertical, t a n πœƒ , when the disk is hanging in its equilibrium position.

  • A 𝑑 = 2 7 c m , t a n πœƒ = 1 4
  • B 𝑑 = 1 1 1 c m , t a n πœƒ = 2 8 9
  • C 𝑑 = 2 7 c m , t a n πœƒ = 4
  • D 𝑑 = 1 1 1 c m , t a n πœƒ = 9 2 8