A steel wheel is mounted on a frictionless bearing as shown in the left figure above. A cross-sectional diagram of the wheel is shown in the right figure below. A 2 m length of the chord is carefully wound around the rum of the wheel. The chord is then pulled with a constant force of 27 N until the chord has completely unwound from the wheel.a) Determine the moment of inertia of the wheel.b) Determine how much work is done to get the wheel spinning (just after the chord is free of the wheel) and how this compares to the final kinetic energy of the spinning wheel?Note that the density, p, of the steel is 7.874 g/cm3. Some hints that might help in working this problem: *Although we’re not told what the mass of the wheel is, we are given the density of the steel out of which it is made. Note that mass the mass of an object is the product of its density and its volume: m=pV.*As can be seen in the cross-sectional diagram above, this is not a simple solid wheel. Recall that I=ΣMnRn^2=ΣIn .*The wheel obviously has a hole in its center so it can be mounted on an axle. However, when the wheel is spinning, the hole is filled with the matter making up the axle. If we assume the axle is made of steel with the same density as the wheel, we can consider it part of the wheel and treat the wheel as if it didn’t have a hole in its center when we’re trying to find its Moment of Inertia.*We might need to find the total angle, delta theta, through which the wheel turns as the chord is being pulled. This might be useful, for example, if we want to use any of the kinematic equations for rotational motion. For this, we can assume theta initial = 0. 9 cm2 cm2.5 cm|I l cm7 cm 11. A steel wheel is mounted on a frictionless bearing as shown in the left figure above. A crosssectional diagram of the wheel is shown in the right figure below. A 2 m length of chord is carefullywound around the rum of the wheel. The chord is then pulled with a constant force of 27 N until thechord has completely unwound from the wheel. a) Determine the moment of inertia of the wheel. b)Determine how much work is done to get the wheel spinning (just after the chord is free of the wheel)and how this compares to the final kinetic energy of the spinning wheel? Note that the density, p, ofthe steel is 7.874 g/cm?.Some hints that might help in working this problem:• Although we're not told the what the mass of the wheel is, we are given the density of the steelout of which it is made. Note that mass the mass of an object is the product of is density andits volume: m= pV.• As can be seen in the cross sectional diagram above, this is not a simple solid wheel. Recall thatΙ-Σm.y=ΣΙ.The wheel obviously has a hole in its center so it can be mounted on an axle. However, whenthe wheel is spinning, the hole is filled with the matter making up the axle. If we assume theaxle is made of steel with the same density as the wheel, we can consider it part of the wheeland treat the wheel as if it didn't have a hole in its center when we're trying to find its Momentof Inertia.We might need to find the total angle, AG, through which the wheel turns as the chord is beingpulled. This might be useful, for example, if we want to use any of the kinematic equations forrotational motion. For this, we can assume 6, = 0.

a) The total moment of inertia will equal to the sum of the moment of inertia of all the parts of…

9 cm 2 cm 2.5 cm |I l cm 7 cm 11. A steel wheel is mounted on a frictionless bearing as shown in the left figure above. A cross sectional diagram of the wheel is shown in the right figure below. A 2 m length of chord is carefully wound around the rum of the wheel. The chord is then pulled with a constant force of 27 N until the chord has completely unwound from the wheel. a) Determine the moment of inertia of the wheel. b) Determine how much work is done to get the wheel spinning (just after the chord is free of the wheel) and how this compares to the final kinetic energy of the spinning wheel? Note that the density, p, of the steel is 7.874 g/cm?. Some hints that might help in working this problem: • Although we're not told the what the mass of the wheel is, we are given the density of the steel out of which it is made. Note that mass the mass of an object is the product of is density and its volume: m= pV. • As can be seen in the cross sectional diagram above, this is not a simple solid wheel. Recall that Ι-Σm.y=ΣΙ. The wheel obviously has a hole in its center so it can be mounted on an axle. However, when the wheel is spinning, the hole is filled with the matter making up the axle. If we assume the axle is made of steel with the same density as the wheel, we can consider it part of the wheel and treat the wheel as if it didn't have a hole in its center when we're trying to find its Moment of Inertia. We might need to find the total angle, AG, through which the wheel turns as the chord is being pulled. This might be useful, for example, if we want to use any of the kinematic equations for rotational motion. For this, we can assume 6, = 0.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A steel wheel is mounted on a frictionless bearing as shown in the left figure above. A cross-sectional diagram of the wheel is shown in the right figure below. A 2 m length of the chord is carefully wound around the rum of the wheel. The chord is then pulled with a constant force of 27 N until the chord has completely unwound from the wheel.

a) Determine the moment of inertia of the wheel.

b) Determine how much work is done to get the wheel spinning (just after the chord is free of the wheel) and how this compares to the final kinetic energy of the spinning wheel?

Note that the density, p, of the steel is 7.874 g/cm3.

Some hints that might help in working this problem:

*Although we’re not told what the mass of the wheel is, we are given the density of the steel out of which it is made. Note that mass the mass of an object is the product of its density and its volume: m=pV.

*As can be seen in the cross-sectional diagram above, this is not a simple solid wheel. Recall that I=ΣMnRn^2=ΣIn .

*The wheel obviously has a hole in its center so it can be mounted on an axle. However, when the wheel is spinning, the hole is filled with the matter making up the axle. If we assume the axle is made of steel with the same density as the wheel, we can consider it part of the wheel and treat the wheel as if it didn’t have a hole in its center when we’re trying to find its Moment of Inertia.

*We might need to find the total angle, delta theta, through which the wheel turns as the chord is being pulled. This might be useful, for example, if we want to use any of the kinematic equations for rotational motion. For this, we can assume theta initial = 0.

11. A steel wheel is mounted on a frictionless bearing as shown in the left figure above. A cross
sectional diagram of the wheel is shown in the right figure below. A 2 m length of chord is carefully
wound around the rum of the wheel. The chord is then pulled with a constant force of 27 N until the
chord has completely unwound from the wheel. a) Determine the moment of inertia of the wheel. b)
Determine how much work is done to get the wheel spinning (just after the chord is free of the wheel)
and how this compares to the final kinetic energy of the spinning wheel? Note that the density, p, of
the steel is 7.874 g/cm?.
Some hints that might help in working this problem:
• Although we're not told the what the mass of the wheel is, we are given the density of the steel
out of which it is made. Note that mass the mass of an object is the product of is density and
its volume: m= pV.
• As can be seen in the cross sectional diagram above, this is not a simple solid wheel. Recall that
Ι-Σm.y=ΣΙ.
The wheel obviously has a hole in its center so it can be mounted on an axle. However, when
the wheel is spinning, the hole is filled with the matter making up the axle. If we assume the
axle is made of steel with the same density as the wheel, we can consider it part of the wheel
and treat the wheel as if it didn't have a hole in its center when we're trying to find its Moment
of Inertia.
We might need to find the total angle, AG, through which the wheel turns as the chord is being
pulled. This might be useful, for example, if we want to use any of the kinematic equations for
rotational motion. For this, we can assume 6, = 0.

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