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A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is?
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A uniform cylinder of mass M lies on a fixed plane inclined at an angl...
Let the radius of cylinder be R For the cylinder to remain static, net torque on cylinder about point P (point of contact with inclined surface) should be zero.
Mg (OS) = mg (SQ)
∴ or Mg R sinq
= mg R (1 – sin q) or
m
= M sin θ / 1 - sin θ
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A uniform cylinder of mass M lies on a fixed plane inclined at an angl...
Analysis:
To determine the value of mass m for the cylinder to remain static, we need to consider the forces acting on the system and apply the conditions for equilibrium.
Forces Acting on the System:
1. Weight of the cylinder (Wcylinder): The weight of the cylinder acts vertically downwards and can be calculated as Wcylinder = M*g, where g is the acceleration due to gravity.
2. Weight of the hanging mass (Whanging): The weight of the hanging mass acts vertically downwards and can be calculated as Whanging = m*g.
3. Normal force (N): The normal force acts perpendicular to the inclined plane and counteracts the weight of the cylinder. It can be calculated as N = Wcylinder*cos(theta).
4. Friction force (f): The friction force acts parallel to the inclined plane and prevents the cylinder from slipping. It can be calculated as f = μ*N, where μ is the coefficient of friction between the cylinder and the plane.
5. Tension in the string (T): The tension in the string acts horizontally and provides the necessary force to balance the torque due to the hanging mass.
Conditions for Equilibrium:
For the cylinder to remain static, the sum of the forces in the x-direction and the sum of the forces in the y-direction must be zero. Additionally, the sum of the torques about any point must also be zero.
Sum of Forces in the x-direction:
The only force in the x-direction is the tension in the string (T). Therefore, T = 0, since there is no acceleration in the x-direction.
Sum of Forces in the y-direction:
The forces in the y-direction are the normal force (N), the weight of the cylinder (Wcylinder), and the weight of the hanging mass (Whanging). Therefore, the sum of forces in the y-direction can be written as N - Wcylinder - Whanging = 0.
Sum of Torques:
The torque about any point on the cylinder due to the hanging mass is given by T*d, where d is the perpendicular distance between the point and the line of action of T. The torque due to the weight of the cylinder is zero since it acts through the center of mass. Therefore, the sum of torques can be written as T*d = 0.
Solving for Mass m:
Given that T = 0 and T*d = 0, we can conclude that the tension in the string and the torque due to the hanging mass are both zero. This implies that mass m does not contribute to any forces or torques acting on the system.
Conclusion:
To keep the cylinder static, the value of mass m can be any positive or zero value. The presence or absence of mass m does not affect the equilibrium of the system as the tension in the string is zero.
To determine the value of mass m for the cylinder to remain static, we need to consider the forces acting on the system and apply the conditions for equilibrium.
Forces Acting on the System:
1. Weight of the cylinder (Wcylinder): The weight of the cylinder acts vertically downwards and can be calculated as Wcylinder = M*g, where g is the acceleration due to gravity.
2. Weight of the hanging mass (Whanging): The weight of the hanging mass acts vertically downwards and can be calculated as Whanging = m*g.
3. Normal force (N): The normal force acts perpendicular to the inclined plane and counteracts the weight of the cylinder. It can be calculated as N = Wcylinder*cos(theta).
4. Friction force (f): The friction force acts parallel to the inclined plane and prevents the cylinder from slipping. It can be calculated as f = μ*N, where μ is the coefficient of friction between the cylinder and the plane.
5. Tension in the string (T): The tension in the string acts horizontally and provides the necessary force to balance the torque due to the hanging mass.
Conditions for Equilibrium:
For the cylinder to remain static, the sum of the forces in the x-direction and the sum of the forces in the y-direction must be zero. Additionally, the sum of the torques about any point must also be zero.
Sum of Forces in the x-direction:
The only force in the x-direction is the tension in the string (T). Therefore, T = 0, since there is no acceleration in the x-direction.
Sum of Forces in the y-direction:
The forces in the y-direction are the normal force (N), the weight of the cylinder (Wcylinder), and the weight of the hanging mass (Whanging). Therefore, the sum of forces in the y-direction can be written as N - Wcylinder - Whanging = 0.
Sum of Torques:
The torque about any point on the cylinder due to the hanging mass is given by T*d, where d is the perpendicular distance between the point and the line of action of T. The torque due to the weight of the cylinder is zero since it acts through the center of mass. Therefore, the sum of torques can be written as T*d = 0.
Solving for Mass m:
Given that T = 0 and T*d = 0, we can conclude that the tension in the string and the torque due to the hanging mass are both zero. This implies that mass m does not contribute to any forces or torques acting on the system.
Conclusion:
To keep the cylinder static, the value of mass m can be any positive or zero value. The presence or absence of mass m does not affect the equilibrium of the system as the tension in the string is zero.
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A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is?
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A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is?.
A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is?.
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A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is?, a detailed solution for A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is? has been provided alongside types of A uniform cylinder of mass M lies on a fixed plane inclined at an angle theta with horizontal. A light string is tied to the cylinder's right most point, and a mass m hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping. For the cylinder to remain static, the value of mass m is? theory, EduRev gives you an
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