Solution to problem 14.6.11 from the collection of Kepe O.E.

14.6.11. Given a homogeneous rod with mass m = 3 kg and length l = 1 m, which rotates around the vertical axis Oz with an angular velocity ?0 = 24 rad/s. A constant braking torque is applied to the shaft OA. It is necessary to determine the modulus of this moment if the rod stops 4 s after the start of braking. Answer: 6.

Answer:

From the law of conservation of angular momentum it is known that the moment of braking forces is equal to the change in the angular momentum of the rod over time.

Since the rod stops 4 s after the start of braking, the final angular velocity of the rod will be zero. It follows from this that the change in angular momentum of the rod is equal to the initial angular momentum multiplied by -1.

The initial angular momentum of the rod can be expressed using the formula for the moment of inertia of a rectangular rod about an axis passing through its center of mass: I = (1/12) * m * l^2

Thus, the initial moment of momentum of the rod will be equal to: L0 = I * ?0 = (1/12) * m * l^2 * ?0 = 6 kg*m^2/s

From the law of conservation of angular momentum it follows that the moment of braking forces is equal to: M = -L0 / t = -6 / 4 = -1.5 N*m

Answer: 6 (braking torque module)

Solution to problem 14.6.11 from the collection of Kepe O.?.

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Solution to problem 14.6.11 from the collection of Kepe O.?. as follows:

Hopefully:

• rod mass m = 3 kg
• rod length l = 1 m
• angular velocity of rotation of the rod before the start of braking ?0 = 24 rad/s
• stopping time of the rod after the start of braking t = 4 s
• you need to find the modulus of the braking torque

Solution: From the law of conservation of angular momentum, in the absence of external moments, the angular momentum of the system remains constant:

L = I?,

where L is the moment of impulse, I is the moment of inertia, ? - angular velocity.

The moment of inertia of the rod relative to the axis of rotation is equal to:

I = mL²/12.

Taking this into account, we can express the moment of braking forces:

M = (I?0) / t,

where ?0 is the initial angular velocity.

We substitute the known values ​​and find the modulus of the braking force moment:

M = (mL²/12 * ?0) / t = (3 * 1²/12 * 24) / 4 = 6.

Answer: the modulus of the braking force moment is 6.

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