a) A particle describes a horizontal circle of radius 0.8 metres with uniform angular velocity ω radians per second - Leaving Cert Applied Maths - Question 8 - 2019
Question 8
a) A particle describes a horizontal circle of radius 0.8 metres with uniform angular velocity ω radians per second. The mass of the particle is 0.4 kg. The particle... show full transcript
Worked Solution & Example Answer:a) A particle describes a horizontal circle of radius 0.8 metres with uniform angular velocity ω radians per second - Leaving Cert Applied Maths - Question 8 - 2019
Step 1
Find (i) the value of ω
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Answer
To find the angular velocity ω, we first convert the particle's revolutions per minute to radians per second. The formula to convert revolutions to radians is:
extRevolutionsimes2π=extRadians
Given that the particle completes 12 revolutions in a minute:
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Answer
The speed of the particle can be found using the formula:
v=rω
Where r is the radius (0.8 m) and ω is the angular velocity we found earlier. Therefore:
v=0.8×52π≈1.006 m/s.
Step 3
Find (iii) the acceleration of the particle
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Answer
The centripetal acceleration a can be calculated using:
a=rω2
Substituting in our values:
a=0.8×(52π)2=1.26 m/s2.
Step 4
Find (iv) the centripetal force on the particle
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Answer
The centripetal force F can be calculated using:
F=ma
Where m = 0.4 kg and a = 1.26 m/s². Thus:
F=0.4×1.26=0.504 N≈0.5 N.
Step 5
Find (i) the tension in the string
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Answer
To find the tension T in the string, we analyze the forces acting on the particle. The vertical component of tension must balance the weight of the particle, and the horizontal component provides the centripetal force. Therefore:
T \sin(θ) = m g\n$$
Substituting r = 0.2 m gives:
T \sin(θ) = 0.5 \times 9.81 \text{ N}\n$$
Given that θ = \tan^{-1}\left( \frac{3}{4} \right), we can find T. First calculate sin(θ):
T \sin(θ) = 5 ext{ N}\n$$
Solving this gives:
T = 3.125 ext{ N}.
Step 6
Find (ii) the reaction force between the particle and the table
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Answer
The vertical components of the forces acting on the particle consist of the normal reaction R and the vertical component of the tension T. Since the particle is not moving vertically, we have:
R + T \cos(θ) = mg\n$$
Substituting the mass and the value of T calculated earlier gives:
R + 3.125 \times \cos(\theta) = 0.5 ext{ kg} imes 9.81 ext{ m/s}^2\n$$
Calculating R provides:
R=3.125extN−3.125extN=0.5extN.
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