การเคลื่อนที่แบบหมุน – ปัญหาและวิธีแก้ไข

การเคลื่อนที่แบบหมุน – ปัญหาและวิธีแก้ไข

แรงบิด

1. A beam 140 cm in length. There are three forces acts on the beam, F₁ = 20 N, F₂ = 10 N, and F₃ = 40 N with direction and position as shown in the figure below. What is the แรงบิด causes the beam rotates about the center of mass of the beam?

เป็นที่รู้จัก :

The center of mass located at the center of the beam.

Length of beam (l) = 140 cm = 1.4 meters

กองกำลัง 1 (F₁) = 20 N, the lever arm 1 (l₁) = 70 cm = 0.7 meters

กองกำลัง 2 (F₂) = 10 N, the lever arm 2 (l₂) = 100 cm – 70 cm = 30 cm = 0.3 meters

กองกำลัง 3 (F₃) = 40 N, the lever arm 3 (l₃) = 70 cm = 0.7 meters

เป็นที่ต้องการ : The magnitude of torque

วิธีการแก้ปัญหา:

The torque 1 rotates beam clockwise, so assigned a negative sign to the torque 1.

τ₁ = ฉ₁ l₁ = (20 N)(0.7 m) = -14 N m

The torque 2 rotates beam counterclockwise, so assigned a positive sign to the torque 2.

τ₂ = ฉ₂ l₂ = (10 N)(0.3 m) = 3 N m

The torque 3 rotates beam clockwise, so assigned a positive sign to the torque 3.

τ₃ = ฉ₃ l3 = (40 N)(0.7 m) = -28 N m

The net torque :

Στ = -14 Nm + 3 Nm – 28 Nm = – 42 Nm + 3 Nm = -39 Nm

The magnitude of the torque is 39 N m. The direction of rotation of the beam clockwise, so assigned a negative sign.

2. What is the net torque acts on the beam The axis of rotation at point D. (sin 53^o = 0.8)

เป็นที่รู้จัก :

The axis of rotation at point D

F₁ = 10 N and l₁ = r₁ sin θ = (40 cm)(sin 53^o) = (0.4 m)(0.8) = 0.32 meters

F₂ = 10√2 N and l₂ = r₂ sin θ = (20 cm)(sin 45^o) = (0.2 m)(0.5√2) = 0.1√2 meters

F₃ = 20 N and l₃ = r₁ sin θ = (10 cm)(sin 90^o) = (0.1 m)(1) = 0.1 meters

เป็นที่ต้องการ : The net torque

วิธีการแก้ปัญหา:

τ₁ = ฉ₁ l₁ = (10 N)(0.32 m) = 3.2 Nm

(The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1)

τ₂ = ฉ₂ l₂ = (10√2 N)( 0.1√2 m) = -2 Nm

(The torque 2 rotates beam clockwise so we assign negative sign to the torque 2)

τ₃ = ฉ₂ l₂ = (20 N)(0.1 m) = 2 Nm

(The torque 3 rotates beam counterclockwise so we assign positive sign to the torque 3)

The net torque :

Στ = τ₁ – ที₁ + τ₃

Στ = 3.2 Nm – 2 Nm + 2 Nm

Στ = 3.2 Nm

3. What is the net torque if the axis of rotation at point D. (sin 53^o = 0.8)

เป็นที่รู้จัก :

The axis of rotation at point D.

ระยะทาง between F₁ and the axis of rotation (r_AD) = 40 ซม. = 0.4 ม.

Distance between F₂ and the axis of rotation (r_BD) = 20 ซม. = 0.2 ม.

Distance between F₃ and the axis of rotation (r_CD) = 10 ซม. = 0.1 ม.

F₁ = 10 นิวตัน

F₂ = 10√2 นิวตัน

F₃ = 20 นิวตัน

ไม่มี 53^o = 0.8

เป็นที่ต้องการ : The net torque

วิธีการแก้ปัญหา:

The moment of the force 1

เซนต์₁ = (เอฟ₁)(r_AD บาป 53^o) = (10 N)(0.4 m)(0.8) = 3.2 N.m

(The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1)

The moment of the force 2

เซนต์₂ = (เอฟ₂)(r_BD บาป 45^o) = (10√2 N)(0.2 m)(0.5√2) = -2 N.m

(The torque 2 rotates beam clockwise so we assign negative sign to the torque 2)

The moment of the force 3

เซนต์₃ = (เอฟ₃)(r_CD บาป 90^o) = (20 N)(0.1 m)(1) = 2 N.m

(The torque 2 rotates beam counterclockwise so we assign positive sign to the torque 3)

The net torque :

Στ = Στ₁ + Στ₂ + Στ₃

Στ = 3.2 – 2 + 2

Στ = 3.2 Newton meter

The moment of inertia

4. Length of wire = 12 m, l₁ = 4 m. Ignore wire’s mass. What is the โมเมนต์ความเฉื่อย ของระบบ

เป็นที่รู้จัก :

มวล of A (m_A) = 0.2 กก.

Mass of B (m_B) = 0.6 กก.

Distance between A and the axis of rotation (r_A) = 4 เมตร

Distance between B and the axis of rotation (r_B) = 12 – 4 = 8 meters

เป็นที่ต้องการ : โมเมนต์ความเฉื่อยของระบบ

วิธีการแก้ปัญหา:

The moment of inertia of A

I_A = (ม._A)(r_A²) = (0.2)(4)² = (0.2)(16) = 3.2 กก. ม.²

The moment of inertia of B

I_B = (ม._B)(r_B²) = (0.6)(8)² = (0.6)(64) = 38.4 กก. ม.²

The moment of inertia of the system :

ฉัน = ฉัน_A + ฉัน_B = 3.2 + 38.4 = 41.6 kg m²

Rotational dynamics

5. A 6-N force is applied to a cord wrapped around a pulley of mass M = 5 kg and radius R = 20 cm. What is the angular acceleration of the pulley. The pulley is a uniform solid cylinder.

เป็นที่รู้จัก :

แรง (F) = 6 นิวตัน

Mass (M) = 5 kg

Radius (R) = 20 cm = 20/100 m = 0.2 m

เป็นที่ต้องการ : ความเร่งเชิงมุม (ก)

วิธีการแก้ปัญหา:

The moment of the force :

τ = F R = (6 Newton)(0.2 meters) = 1.2 N m

The moment of inertia for solid cylinder :

I = 1/2 M R²

I = 1/2 (5 kg)(0.2 m)²

I = 1/2 (5 kg)(0.04 m²)

I = 1/2 (0.2)

I = 0.1 กก. ม.².

The angular acceleration :

τ = ฉัน α

α = τ / I = 1.2 / 0.1 = 12 rad s^-2

6. A block of mass = 4 kg hanging from a cord wrapped around a pulley of mass = 8 kg and radius R = 10 cm. Acceleration due to gravity is 10 ms^-2 . What is the linear acceleration of the block? The pulley is a uniform solid cylinder.

เป็นที่รู้จัก :

Mass of pulley (m) = 8 kg

Radius of pulley (r) = 10 cm = 0.1 m

มวลของบล็อก (เมตร) = 4 กิโลกรัม

ความเร่งเนื่องจากแรงโน้มถ่วง (g) = 10 m/s²²

น้ำหนัก (w) = mg = (4 kg)(10 m/s)²) = 40 กก. ม./วินาที² = 40 นิวตัน

เป็นที่ต้องการ : The free fall acceleration of the block

วิธีการแก้ปัญหา:

The moment of inertia of the solid cylinder :

I = 1/2 M R² = 1/2 (8 kg)(0.1 m)² = (4 kg)(0.01 m²) = 0.04 kg m²

The moment of the force :

τ = F r = (40 N)(0.1 m) = 4 Nm

The angular acceleration :

Στ = I α

4 = 0.04 α

α = 4 / 0.04 = 100

The linear acceleration :

a = r α = (0.1)(100) = 10 m/s²

7. A block with mass of m hanging from a cord wrapped around a pulley. If the ตกอย่างอิสระ acceleration of the block is a m/s², what is the moment of inertia of the pulley..

เป็นที่รู้จัก :

weight = w = m g

Lever arm = R

The angular acceleration = α

The free fall acceleration of the block = a ms^-2

ต้องการ: The moment of inertia of the pulley (I)

วิธีการแก้ปัญหา:

The connection between the linear acceleration and the angular acceleration :

a = R α

α = a / R

The moment of inertia :

τ = ฉัน α

I = τ : α = τ : a / R = τ (R / a) = τ R a^-1

The angular momentum

8. A 0.2-gram particle moves in a circle at a constant speed of 10 m/s. The radius of the circle is 3 cm. What is the angular momentum of the particle?

เป็นที่รู้จัก :

Mass of particle (m) = 0.2 gram = 2 x 10^-4 kg

Angular speed (ω) = 10 rad s^-1

Radius (r) = 3 cm = 3 x 10^-2 เมตร

เป็นที่ต้องการ : The angular momentum of the particle

วิธีการแก้ปัญหา:

The equation of the angular momentum :

L = I ω

I = the angular momentum, I = the moment of inertia, ω = the angular speed

The moment of inertia (for particle) :

ฉัน = นาย² = (2 x 10 .)^-4 )(3 x 10^-2)² = (2 x 10 .)^-4 )(9 x 10^-4) = 18 x 10^-8

The angular momentum :

L = I ω = (18 x 10^-8)(10 rad s^-1) = 18 x 10^-7 กก. ม.² s^-1

1. What is rotational motion?
   • คำตอบ: Rotational motion refers to the movement of an object around a fixed axis. It’s the kind of motion in which every point of the object moves in a circle about the axis.
2. How does linear velocity relate to angular velocity in rotational motion?
   • คำตอบ: Linear velocity ($v$) of a point in a rotating object is directly proportional to its distance ($r$) from the axis of rotation and the angular velocity ($\omega$) of the object. The relation is given by $v=r\cdot \omega$.
3. What is the moment of inertia, and how does it relate to rotational motion?
   • คำตอบ: Moment of inertia is the rotational analog of mass in linear motion. It measures an object’s resistance to changes in its rotational state. The moment of inertia depends on both the mass of an object and its distribution relative to the axis of rotation.
4. How does Newton’s first law of motion apply to rotational motion?
   • คำตอบ: Just as an object in linear motion remains in motion unless acted upon by an external force, an object in rotational motion will remain in that state unless acted upon by an external torque.
5. What is the significance of the radius of gyration?
   • คำตอบ: The radius of gyration provides a measure of the distribution of an object’s mass away from its axis of rotation. It essentially describes how far from the axis all of the object’s mass would need to be concentrated to have the same moment of inertia as the original distribution.
6. What is angular momentum and how is it conserved?
   • คำตอบ: Angular momentum is the rotational equivalent of linear momentum. It is the product of an object’s moment of inertia and its angular velocity. In a closed system, the total angular momentum remains constant unless acted upon by an external torque, highlighting the conservation of angular momentum.
7. How does torque influence rotational motion?
   • คำตอบ: Torque is the rotational equivalent of force. It causes changes in the rotational motion of an object. The relation is given by Newton’s second law for rotation: $\tau =I\cdot \alpha$ที่นี่มี $\tau$ is the torque, $I$ คือโมเมนต์ความเฉื่อย และ $\alpha$ is the angular acceleration.
8. How does the center of mass differ from the center of rotation?
   • คำตอบ: While they can coincide, the center of mass is the point where the entire mass of an object can be assumed to be concentrated for the purpose of calculations in linear motion, whereas the center of rotation is the point (or axis) about which an object rotates.
9. What is the role of the centripetal force in rotational motion?
   • คำตอบ: Centripetal force is the net force acting on an object moving in a circular path, directed towards the center of rotation. It’s responsible for keeping an object in its curved path and preventing it from moving in a straight line due to inertia.

10. How is rotational kinetic energy related to the moment of inertia and angular velocity?

    • คำตอบ: Rotational kinetic energy is the energy due to the rotation of an object about an axis. It is given by the formula: KE_{rotational†<}=1/2 †<I ω²ที่นี่มี $I$ คือโมเมนต์ความเฉื่อยและ $\omega$ is the angular velocity.