Question

A uniform rigid rod with mass Mr = 7.1 kg, length L = 1.7 m rotates in the vertical xy plane about a frictionless pivot through its center. Two point-like particles m1 and m2, with masses m1 = 4.1 kg and m2 = 3.0 kg, are attached at the ends of the rod. What is the magnitude of the angular acceleration of the system when the rod makes an angle of 45.9° with the horizontal?

Answer 1

Answer:

$\alpha = 0.93 rad/s^2$

Explanation:

When rod makes an angle of 45.9 degree with the horizontal then net torque on the rod due to two masses at the ends is given as

$\tau = m_1g(\frac{L}{2}cos45.9) - m_2g(\frac{L}{2}cos45.9)$

$\tau = (m_1 - m_2)g\frac{L}{2} cos45.9$

$\tau = (4.1 - 3)(9.81)(\frac{1.7}{2})cos45.9$

$\tau = 6.38Nm$

now moment of inertia of the system is given as

$I = \frac{ML^2}{12} + (m_1 + m_2)(\frac{L}{2})^2$

$I = \frac{7.1(1.7)^2}{12} + (4.1 + 3)(\frac{1.7^2}{4})$

$I = 1.71 + 5.13$

$I = 6.84 kg m^2$

now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6.38}{6.84}$

$\alpha = 0.93 rad/s^2$

Answer 2

The angular acceleration of the system when the uniform right rod makes an angle of 45.9° with the horizontal is 0.93 rad/s².

What is angular acceleration of a body?

The angular acceleration of a body is the rate by which the body changed its speed with respect to the time. The angular acceleration in terms of torque and inertia can be given as,

$\omega= \dfrac{\tau}{I}$

Here, (τ) is the torque, and (I) is the inertia.

The magnitude of the torque exerted at the center of the rod by gravity can be given as,

$\tau=\dfrac{MgLcos\theta}{2}$

Here, (M) is the mass, (g) is the gravitation force and (L) is the length.

A uniform rigid rod with mass Mr = 7.1 kg, length L = 1.7 m rotates in the vertical xy plane about a frictionless pivot through its center.

Two point-like particles m1 and m2, with masses m1 = 4.1 kg and m2 = 3.0 kg, are attached at the ends of the rod.

The rod makes an angle of 45.9° with the horizontal. Thus, the net torque for this case using the above formula can be given as,'

$\tau=\dfrac{(4.1)(9.81)(1.7)cos(45.9)}{2}-\dfrac{(3)(9.81)(1.7)cos(45.9)}{2}\\\tau=6.84\rm Nm$

The moment of inertia of this system can be given as,

$I=\dfrac{ML^2}{12}+(M_1+M_2)\dfrac{L^2}{4}\\I=\dfrac{(7.1)(1.7)^2}{12}+(4.1+3)\dfrac{(1.7)^2}{4}\\I=6.84\rm kgm^2$

The angular acceleration of the system is,

$\omega= \dfrac{\tau}{I}\\\omega= \dfrac{6.38}{6.84}\\\omega=0.93\rm \;rad/s^2$

Hence, the angular acceleration of the system when the uniform right rod makes an angle of 45.9° with the horizontal is 0.93 rad/s².

Learn more about the angular acceleration here;

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