1. Two point masses, m, = 1 kg and m, = 2 kg, are suspended by a massless rod of total length (= 1 m. The rod is pivoted a distance d = 0.25 m from the left end as shown. (a) What is the moment of inertia of the system about the pivot point? [Hint: the rotational inertia of a single point mass is Imar = mr², where r is the distance from the mass to the axis of rotation. To find the total rotational inertia of a system, you simply add the individual rotational inertias. This is just as you would find total mass of a system by adding the individual masses.] (b) Draw a free body force diagram for the rod. Remember that it is now important to draw the forces on the body where they actually act. (c) Convince yourself (and the others in yóur group) that the torque about the pivot point due to gravity acting on m, is out of the page, while that due to gravity acting on m, is into the page. Find numerically the torque due to each, taking g = 9.8 m/s. (d) Find the net torque about the pivot point at the instant shown. (e) Find the angular acceleration about `the pivot point at the instant shown. (H The rod is released from rest from the position shown. Through what angle does the rod rotate in the next 0.001 s? Assume that the angular acceleration is constant through this short time interval. (Ans: (a) Ip = 1.19 kg m²; (d) Et,= 12.25 N m, in which direction?; (e) a, = 10.3 rad/s², in which direction?: (f) 5.2 x 1o*radi %3D
1. Two point masses, m, = 1 kg and m, = 2 kg, are suspended by a massless rod of total length (= 1 m. The rod is pivoted a distance d = 0.25 m from the left end as shown. (a) What is the moment of inertia of the system about the pivot point? [Hint: the rotational inertia of a single point mass is Imar = mr², where r is the distance from the mass to the axis of rotation. To find the total rotational inertia of a system, you simply add the individual rotational inertias. This is just as you would find total mass of a system by adding the individual masses.] (b) Draw a free body force diagram for the rod. Remember that it is now important to draw the forces on the body where they actually act. (c) Convince yourself (and the others in yóur group) that the torque about the pivot point due to gravity acting on m, is out of the page, while that due to gravity acting on m, is into the page. Find numerically the torque due to each, taking g = 9.8 m/s. (d) Find the net torque about the pivot point at the instant shown. (e) Find the angular acceleration about `the pivot point at the instant shown. (H The rod is released from rest from the position shown. Through what angle does the rod rotate in the next 0.001 s? Assume that the angular acceleration is constant through this short time interval. (Ans: (a) Ip = 1.19 kg m²; (d) Et,= 12.25 N m, in which direction?; (e) a, = 10.3 rad/s², in which direction?: (f) 5.2 x 1o*radi %3D
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![**Topic: Physics - Rotational Motion and Torque**
**Exercise:**
1. **Problem Description:**
Two point masses, \( m_1 = 1 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \), are suspended by a massless rod of total length \( \ell = 1 \, \text{m} \). The rod is pivoted a distance \( d = 0.25 \, \text{m} \) from the left end as shown.
(a) What is the moment of inertia of the system about the pivot point? [Hint: the rotational inertia of a single point mass is \( I_{\text{point mass}} = mr^2 \), where \( r \) is the distance from the mass to the axis of rotation. To find the total rotational inertia of a system, you simply add the individual rotational inertias. This is just as you would find total mass of a system by adding the individual masses.]
(b) Draw a free body force diagram for the rod. Remember that it is now important to draw the forces on the body where they actually act.
(c) Convince yourself (and the others in your group) that the torque about the pivot point due to gravity acting on \( m_1 \) is out of the page, while that due to gravity acting on \( m_2 \) is into the page. Find numerically the torque due to each, taking \( g = 9.8 \, \text{m/s}^2 \).
(d) Find the net torque about the pivot point at the instant shown.
(e) Find the angular acceleration about the pivot point at the instant shown.
(f) The rod is released from rest from the position shown. Through what angle does the rod rotate in the next 0.001 s? Assume that the angular acceleration is constant through this short time interval.
**Answers:**
(a) \( I_p = 1.19 \, \text{kg m}^2 \)
(d) \( \Sigma \tau_p = 12.25 \, \text{N m, in which direction?} \)
(e) \( \alpha_p = 10.3 \, \text{rad/s}^2, \text{in which direction](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50a0a1ab-f671-42c0-bd3d-0403645711b0%2F8a72ce7c-14f9-4e5c-821a-c41a9a2f04d9%2Fgjcc8an_processed.png&w=3840&q=75)
Transcribed Image Text:**Topic: Physics - Rotational Motion and Torque**
**Exercise:**
1. **Problem Description:**
Two point masses, \( m_1 = 1 \, \text{kg} \) and \( m_2 = 2 \, \text{kg} \), are suspended by a massless rod of total length \( \ell = 1 \, \text{m} \). The rod is pivoted a distance \( d = 0.25 \, \text{m} \) from the left end as shown.
(a) What is the moment of inertia of the system about the pivot point? [Hint: the rotational inertia of a single point mass is \( I_{\text{point mass}} = mr^2 \), where \( r \) is the distance from the mass to the axis of rotation. To find the total rotational inertia of a system, you simply add the individual rotational inertias. This is just as you would find total mass of a system by adding the individual masses.]
(b) Draw a free body force diagram for the rod. Remember that it is now important to draw the forces on the body where they actually act.
(c) Convince yourself (and the others in your group) that the torque about the pivot point due to gravity acting on \( m_1 \) is out of the page, while that due to gravity acting on \( m_2 \) is into the page. Find numerically the torque due to each, taking \( g = 9.8 \, \text{m/s}^2 \).
(d) Find the net torque about the pivot point at the instant shown.
(e) Find the angular acceleration about the pivot point at the instant shown.
(f) The rod is released from rest from the position shown. Through what angle does the rod rotate in the next 0.001 s? Assume that the angular acceleration is constant through this short time interval.
**Answers:**
(a) \( I_p = 1.19 \, \text{kg m}^2 \)
(d) \( \Sigma \tau_p = 12.25 \, \text{N m, in which direction?} \)
(e) \( \alpha_p = 10.3 \, \text{rad/s}^2, \text{in which direction
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