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2 months ago

What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical?

Physics

2 answers:

Keith_Richards [3.2K]2 months ago

8 0

When the pencil is positioned at an angle of 10.0 degrees to the vertical, its angular acceleration is 1306.6 rad / s^2

Clarification:

Imagine balancing a pencil on its tip. If you let it fall, what is the angular acceleration when it is at a 10.0-degree angle from the vertical? Express your answer in radians per second squared.

When the pencil is in balance, there is no net torque; however, once released, the weight of the pencil creates a net torque.

We fundamentally understand the connection between torque and angular acceleration. Thus, we find that the angular acceleration of the pencil is established.

The pencil's length L is 15 cm and its weight (m) corresponds to the angle (θ) of 10° from vertical. Hence, the torque (τ) caused by its weight (W) is articulated as:

$\tau = r xF\\\tau =\frac{L}{2} W sin \theta\\\tau =\frac{L}{2} mg sin 10^o$

$\tau =\frac{0.15}{2} m*10* sin 10^o\\\tau = 0.075 m*10* sin 10^o\\\tau = 0.130m$

From there, we can calculate the angular acceleration:

$\tau = I \alpha = F r\\\tau = 0.130m = F r\\\alpha = \frac{0.098}{0.000075} = 1306.6 rad / s^2$

To delve deeper into angular acceleration,

Sav [3.1K]2 months ago

6 0

<span> </span><span>When the net torque and moment of inertia are given, calculating becomes straightforward.

Using the equation torque = I * alpha, where I represents the moment of inertia and alpha is the angular acceleration.

Consequently, 0.098 / 0.000075 results in 1306.666... rad / s^2

While the angular acceleration stays the same, you can also determine the angular velocity at that instance, which is 21.36 rad / s.</span>

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