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1. M, a solid cylinder (M=1.83 kg, R=0.115 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.630 kg mass, i.e., F = 6.180 N. (the angular acceleration of the cylinder is [58.73rad/s2]) 2. If instead of the force F an actual mass m = 0.630 kg is hung from the string, find the angular acceleration of the cylinder. Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion.

1. M, a solid cylinder (M=1.83 kg, R=0.115 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.630 kg mass, i.e., F = 6.180 N. (the angular acceleration of the cylinder is [58.73rad/s2]) 2. If instead of the force F an actual mass m = 0.630 kg is hung from the string, find the angular acceleration of the cylinder. Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion.

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1. M, a solid cylinder (M=1.83 kg, R=0.115 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.630 kg mass, i.e., F = 6.180 N. (the angular acceleration of the cylinder is [58.73rad/s2])

2. If instead of the force F an actual mass m = 0.630 kg is hung from the string, find the angular acceleration of the cylinder.

Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion.

The answer to the first part of this question was 58.73rad/s2, I just don't know how I am supposed to set up the second part of this question at all. I know that the acceleration will be lesss but I just don't know how to modifiy the equation to account for the hanging mass instead of a constant force

In the image, we have a diagram illustrating a physics problem involving rotational motion and dynamics. **Description:** - **Pulley System:** - There is a solid disk pulley represented by a large circle. This disk has a uniform mass distribution. - The mass of the pulley is labeled as \( M \). - The radius of the pulley is indicated as \( R \). - **Pulley Axle:** - The center of the pulley has a smaller circle representing the axle, implying that the pulley can rotate around this axle. - **Attached Mass:** - A smaller mass, labeled \( m \), is connected to the pulley via a string or rope. This mass hangs vertically downward. - The string is wrapped around the pulley, indicating that it can move or rotate the pulley when it falls. **Explanation for Educational Context:** - This setup is typically used to analyze mechanical systems involving torques and angular momentum. - The reason for using a pulley with a given mass and radius is to explore how rotational inertia (moment of inertia) affects the motion of the system. - The mass \( m \) hanging freely will exert a tangential force on the pulley, causing it to rotate. - The relationship between the linear motion of the mass \( m \) and the rotational motion of the pulley can be described using Newton’s second law for rotation: \( \tau = I\alpha \), where \( \tau \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. This diagram can be used to derive equations of motion for the system and can be integrated into lessons on rotational dynamics, conservation of energy, and the application of Newton's laws to rotational systems.
Transcribed Image Text:In the image, we have a diagram illustrating a physics problem involving rotational motion and dynamics. **Description:** - **Pulley System:** - There is a solid disk pulley represented by a large circle. This disk has a uniform mass distribution. - The mass of the pulley is labeled as \( M \). - The radius of the pulley is indicated as \( R \). - **Pulley Axle:** - The center of the pulley has a smaller circle representing the axle, implying that the pulley can rotate around this axle. - **Attached Mass:** - A smaller mass, labeled \( m \), is connected to the pulley via a string or rope. This mass hangs vertically downward. - The string is wrapped around the pulley, indicating that it can move or rotate the pulley when it falls. **Explanation for Educational Context:** - This setup is typically used to analyze mechanical systems involving torques and angular momentum. - The reason for using a pulley with a given mass and radius is to explore how rotational inertia (moment of inertia) affects the motion of the system. - The mass \( m \) hanging freely will exert a tangential force on the pulley, causing it to rotate. - The relationship between the linear motion of the mass \( m \) and the rotational motion of the pulley can be described using Newton’s second law for rotation: \( \tau = I\alpha \), where \( \tau \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. This diagram can be used to derive equations of motion for the system and can be integrated into lessons on rotational dynamics, conservation of energy, and the application of Newton's laws to rotational systems.
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