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Treating the ladder as a uniform rod of length 3.16 m and mass 11.5 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.213 rad/s^2.

User Vadim
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Final answer:

The question is about calculating the torque needed to give a ladder a specified angular acceleration, using the formula \(\tau = I\alpha\), where \(I\) is the moment of inertia for a uniform rod.

Step-by-step explanation:

The question asks about calculating the torque required to give a ladder a specific angular acceleration, assuming the ladder is a uniform rod.

To find this torque, we can use 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. For a uniform rod, the moment of inertia is \(I = \frac{1}{3}mL^2\), where \(m\) is the mass and \(L\) is the length of the rod.

Using the given values for the mass (11.5 kg), length (3.16 m), and angular acceleration (0.213 rad/s^2), the torque would be calculated as follows:

To calculate the torque the person must exert on the ladder, we can use the formula:

Torque = Moment of Inertia x Angular Acceleration

The moment of inertia of a uniform rod rotating about an axis through its center is given by:

Moment of Inertia = 1/3 x Mass x Length^2

Plugging in the given values, we can calculate the torque:

Torque = (1/3 x 11.5 kg x (3.16 m)^2) x 0.213 rad/s^2 = 7.07 Nm

User Agibsen
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