An object with a mass of #200 kg# is hanging from an axle with a radius of #8 cm#. If the wheel attached to the axle has a radius of #44 cm#, how much work would it take to turn the wheel a length equal to the circumference of the axle?
The work is
The load L=(200g)N#
Radius of axle Radius of wheel Taking moments about the center of the axle The force is The work is The distance is The work is
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To calculate the work required to turn the wheel a length equal to the circumference of the axle, we first need to find the circumference of the axle. The circumference (C) of a circle is given by the formula (C = 2\pi r), where (r) is the radius of the circle.
Given that the radius of the axle is 8 cm, we can calculate the circumference of the axle as follows:
[C_{\text{axle}} = 2\pi \times 8 \text{ cm} ]
[C_{\text{axle}} = 16\pi \text{ cm}]
Next, we need to find the work required to turn the wheel a distance equal to the circumference of the axle. The work done in rotating an object is given by the formula (W = \tau \times \theta), where (W) is the work done, (\tau) is the torque applied, and (\theta) is the angle through which the object is rotated.
In this case, the torque (\tau) can be calculated as the force applied multiplied by the distance from the axis of rotation. Since the force applied is the weight of the object hanging from the axle, and the distance from the axis of rotation is the radius of the wheel, we have:
[\tau = F \times r_{\text{wheel}}]
[= mg \times r_{\text{wheel}}]
where (m) is the mass of the object, (g) is the acceleration due to gravity, and (r_{\text{wheel}}) is the radius of the wheel.
Given that the mass of the object is 200 kg, the acceleration due to gravity is (9.8 , \text{m/s}^2), and the radius of the wheel is 44 cm, we can calculate the torque (\tau) as follows:
[\tau = 200 , \text{kg} \times 9.8 , \text{m/s}^2 \times 0.44 , \text{m} ]
[= 8624 , \text{N}\cdot\text{m}]
Finally, we can find the work (W) required to turn the wheel a length equal to the circumference of the axle by multiplying the torque by the angle through which the wheel is rotated. Since the angle through which the wheel is rotated when its circumference is traversed is (360^\circ) (or (2\pi) radians), we have:
[W = \tau \times \theta]
[= 8624 , \text{N}\cdot\text{m} \times 2\pi , \text{rad}]
[= 54155.29 , \text{J}]
Therefore, it would take approximately (54155.29 , \text{J}) of work to turn the wheel a length equal to the circumference of the axle.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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