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QUESTION ONE (34 MARKS) A) Match the Follow Ing Quantities of Linear Motion with the Respective Quantities of Angular Mokion (2 Marka)

Вопрос

QUESTION ONE (34 MARKS) a) Match the follow ing quantities of linear motion with the respective quantities of angular mokion (2 Marka) is Foecie a. Main b) A fores of F is applied an angle of 30^circ to tangent of a ply wheel of radius 30 cm. If the torquer on the ply wheel 36 N Find F (4 Marks) c) There masses of mass 80g, 60g and 90 g are placed at the vertices of an equilateral triangle of whes 5 m the system rotates about an axis perpendicular to the plane of the triangle and through the centre of mass. Find the moment of inertia (8 Marks)

Решения

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Ответ

a) Match the following quantities of linear motion with the respective quantities of angular motion:- Linear acceleration: Angular acceleration- Linear velocity: Angular velocity- Linear displacement: Angular displacement- Linear momentum: Angular momentumb) A force of F is applied at an angle of 30° to the tangent of a ply wheel of radius 30 cm. If the torque on the ply wheel is 36 N, find F.To find the force F, we can use the formula for torque:Torque = Force × Distance × sin(θ)Where:- Torque is the rotational force acting on an object.- Force is the applied force.- Distance is the distance from the point of rotation to the line of action of the force.- θ is the angle between the force and the distance.Given:- Torque = 36 N- Distance = 30 cm = 0.3 m- θ = 30°Substituting the values in the formula:36 N = F × 0.3 m × sin(30°)Solving for F:F = 36 N / (0.3 m × sin(30°))F = 36 N / (0.3 m × 0.5)F = 36 N / 0.15 NF = 240 NTherefore, the force F is 240 N.c) There are masses of mass 80g, 60g, and 90g placed at the vertices of an equilateral triangle of sides 5 m. The system rotates about an axis perpendicular to the plane of the triangle and through the center of mass. Find the moment of inertia.To find the moment of inertia, we can use the formula for the moment of inertia of a point mass:Moment of Inertia = m × r^2Where:- m is the mass of the object.- r is the distance of the object from the axis of rotation.Given:- Masses: 80g, 60g, and 90g- Distance from the center of mass: 5 m / 3 = 1.67 mCalculating the moment of inertia for each mass:Moment of Inertia for 80g = 80g × (1.67 m)^2 = 0.224 kg·m^2Moment of Inertia for 60g = 60g × (1.67 m)^2 = 0.168 kg·m^2Moment of Inertia for 90g = 90g × (1.67 m)^2 = 0.288 kg·m^2Summing up the moments of inertia:Total Moment of Inertia = 0.224 kg·m^2 + 0.168 kg·m^2 + 0.288 kg·m^2 = 0.68 kg·m^2Therefore, the moment of inertia of the system is 0.68 kg·m^2.