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Lesson Worksheet: The Negative Mass Method Mathematics

In this worksheet, we will practice finding the center of gravity of a lamina that contains holes using the negative mass method.

Q1:

The table shows the distribution of a system of masses on a uniform lamina.

Mass of the Lamina 𝑚Mass Added 𝑚Mass Removed 𝑚
𝑥-axis𝑥𝑥𝑥
𝑦-axis𝑦𝑦𝑦

Which of the following is the center of mass of this system?

  • A𝑚𝑥𝑚𝑥+𝑚𝑥𝑚𝑚+𝑚,𝑚𝑦𝑚𝑦+𝑚𝑦𝑚𝑚+𝑚
  • B𝑚𝑥+𝑚𝑥𝑚𝑥𝑚+𝑚𝑚,𝑚𝑦+𝑚𝑦𝑚𝑦𝑚+𝑚𝑚
  • C𝑚𝑥+𝑚𝑥+𝑚𝑥𝑚+𝑚+𝑚,𝑚𝑦+𝑚𝑦+𝑚𝑦𝑚+𝑚+𝑚
  • D𝑚𝑦+𝑚𝑦𝑚𝑦𝑚+𝑚𝑚,𝑚𝑥+𝑚𝑥𝑚𝑥𝑚+𝑚𝑚
  • E𝑚𝑦𝑚𝑦+𝑚𝑦𝑚𝑚+𝑚,𝑚𝑥𝑚𝑥+𝑚𝑥𝑚𝑚+𝑚

Q2:

A uniform lamina of mass 15 kg has its center of mass at (2,6). If a piece of the lamina is cut out whose mass is 11 kg and whose center of mass is at (6,2), find the coordinates of the center of mass of the remaining part.

  • A4813,5613
  • B(9,17)
  • C(17,9)
  • D5613,4813
  • E(9,17)

Q3:

An equilateral triangle 𝐴𝐵𝐶 has a side length of 82 cm. When three equal masses are placed at the vertices of the triangle, the centre of mass of the system is 𝐺. When the mass at vertex 𝐶 is removed, the centre of mass of the system is 𝐺. Find the coordinates of the centre of mass of the two systems 𝐺 and 𝐺.

  • A𝐺41,8233, 𝐺412,8233
  • B𝐺41,8233, 𝐺412,4132
  • C𝐺41,4133, 𝐺412,4132
  • D𝐺41,4133, 𝐺412,8233

Q4:

Find the coordinates of the center of gravity of the following figure, which is drawn on a grid of unit squares.

  • A(9,8)
  • B18922,8411
  • C27922,4
  • D92,12411
  • E92,4

Q5:

A square-shaped uniform lamina 𝐴𝐵𝐶𝐷 has a side length of 28 cm. A circular disk of radius 7 cm was cut out of the lamina such that its center was at a distance of 17 cm from both 𝐴𝐵 and 𝐵𝐶. Determine the coordinates of the center of mass of the resulting lamina. Take 𝜋=227.

  • A22115,19915
  • B597112,458
  • C454,59756
  • D19930,22130

Q6:

The diagram shows a uniform lamina 𝐴𝐵𝐶 from which a triangle 𝐺𝐵𝐶 has been cut out. 𝐴𝐵𝐶 was an equilateral triangle with a side length of 93 cm and center of mass 𝐺. Find the coordinates of the new center of mass of the resulting lamina. Round your answer to two decimal places if necessary.

  • A(46.5,35.8)
  • B(46.5,33.56)
  • C(33.56,46.5)
  • D(35.8,46.5)

Q7:

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

  • A625 cm
  • B1825 cm
  • C925 cm
  • D3625 cm

Q8:

A uniform lamina is in the form of a rectangle 𝐴𝐵𝐶𝐷, where 𝐴𝐵=56cm and 𝐵𝐶=35cm. Two points 𝐸 and 𝐹 are on 𝐴𝐵 such that 𝐴𝐸=𝐵𝐹=14cm. The triangle 𝑀𝐸𝐹, where 𝑀 is the center of the rectangle, is cut out of the lamina. Find the coordinates of the center of mass of the resulting lamina. Given that the lamina was freely suspended from 𝐷, find the tangent of the angle that 𝐷𝐴 makes to the vertical, tan𝜃, when the lamina is hanging in its equilibrium position.

  • A28,956, tan𝜃=16895
  • B956,28, tan𝜃=16895
  • C28,956, tan𝜃=95168
  • D956,28, tan𝜃=95168

Q9:

The figure shows a uniform square lamina of side length 18 cm. It is divided into nine congruent squares as shown. Given that the square 𝐶 was cut out and stuck onto square 𝐴, determine the coordinates of the center of gravity of the resulting lamina.

  • A6910,8110
  • B233,9
  • C253,9
  • D698,818

Q10:

A uniform square-shaped lamina 𝐴𝐵𝐶𝐷 of side length 222 cm has a mass of one kilogram. The midpoints of 𝐴𝐷, 𝐴𝐵, and 𝐵𝐶 are denoted by 𝑇, 𝑁, and 𝐾 respectively. The corners 𝑇𝐴𝑁 and 𝑁𝐵𝐾 were folded over so that they lie flat on the surface of the lamina. Bodies of masses 365 g and 294 g were attached to the points 𝑇 and 𝐾 respectively. Find the coordinates of the center of mass of the system rounding your answer to two decimal places if necessary.

  • A(144.61,111.83)
  • B(161.18,106.25)
  • C(105.42,134.17)
  • D(105.42,115.75)

This lesson includes 35 additional questions and 258 additional question variations for subscribers.

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