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Sit for year. TO Stårt let my decide using it again. The tube is replaled. 4 has an outer radius weight Oag the in the air C no contact wI the gounus it rotates with a Speed of 4.6 rad lsec , and it t'akes 3 Seconds to stop once I press the brukes. caliulate the magnitude of force' of friction being exerted by the brukes? make the aSsumption that the brakes apply a force halt 4 35 cm. The inner vadius is 30 cm. The tive is 1.8 Kq. while spinning the tire between the inner and outer raidius ofre. way a thickness af about the tire as a hoop WI the difference between inner + outer radlus) and all the mass is inside the hoop Cignore Spokesl axie). For M.oment a inertia use : I tiRe= Ž M crutert r? NOTE: Think TIRE

Sit for year. TO Stårt let my decide using it again. The tube is replaled. 4 has an outer radius weight Oag the in the air C no contact wI the gounus it rotates with a Speed of 4.6 rad lsec , and it t'akes 3 Seconds to stop once I press the brukes. caliulate the magnitude of force' of friction being exerted by the brukes? make the aSsumption that the brakes apply a force halt 4 35 cm. The inner vadius is 30 cm. The tive is 1.8 Kq. while spinning the tire between the inner and outer raidius ofre. way a thickness af about the tire as a hoop WI the difference between inner + outer radlus) and all the mass is inside the hoop Cignore Spokesl axie). For M.oment a inertia use : I tiRe= Ž M crutert r? NOTE: Think TIRE

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**Physics Topic: Calculating the Magnitude of Force of Friction in a Spinning Bike Tire** **Scenario:** I let my bike sit for a year. I decided to start using it again. The tube is replaced. It has an outer radius of 35 cm. The inner radius is 30 cm. The weight of the tire is 1.8 kg. While spinning the tire in the air (no contact with the ground), it rotates with a speed of 17.6 rad/sec, and it takes 3 seconds to stop once I press the brakes. Calculate the magnitude of force of friction being exerted by the brakes? Make the assumption that the brakes apply a force halfway between the inner and outer radius of the tire. **Note:** Think about the tire as a hoop with a thickness of the difference between inner and outer radius and all the mass is inside the hoop (ignore spokes/axle). For moment of inertia use: \[ I_{\text{tire}} = \frac{1}{2}M(r_{\text{outer}}^2 + r_{\text{inner}}^2) \] **Diagram:** The diagram shows a cross-sectional view of the tire with the inner and outer radii labeled. The outer radius extends to 35 cm, and the inner radius is 30 cm. --- ### Explanation of Diagram The diagram displays a circular representation of the tire's cross-section: 1. **Outer Radius (r_outer)**: This is the distance from the center of the tire to the outer edge, labeled as 35 cm. 2. **Inner Radius (r_inner)**: This is the distance from the center of the tire to the inner edge, labeled as 30 cm. The calculation of the force of friction will consider the average radius when the brakes apply force, which is halfway between the inner and outer radius. --- ### Steps to Calculate the Magnitude of Force of Friction: 1. **Moment of Inertia (I):** \[ I_{\text{tire}} = \frac{1}{2} M (r_{\text{outer}}^2 + r_{\text{inner}}^2) \] Given: - \( M = 1.8 \) kg - \( r_{\text{outer}} = 0.35 \) m - \( r_{\text{inner}} = 0.30 \)
Transcribed Image Text:**Physics Topic: Calculating the Magnitude of Force of Friction in a Spinning Bike Tire** **Scenario:** I let my bike sit for a year. I decided to start using it again. The tube is replaced. It has an outer radius of 35 cm. The inner radius is 30 cm. The weight of the tire is 1.8 kg. While spinning the tire in the air (no contact with the ground), it rotates with a speed of 17.6 rad/sec, and it takes 3 seconds to stop once I press the brakes. Calculate the magnitude of force of friction being exerted by the brakes? Make the assumption that the brakes apply a force halfway between the inner and outer radius of the tire. **Note:** Think about the tire as a hoop with a thickness of the difference between inner and outer radius and all the mass is inside the hoop (ignore spokes/axle). For moment of inertia use: \[ I_{\text{tire}} = \frac{1}{2}M(r_{\text{outer}}^2 + r_{\text{inner}}^2) \] **Diagram:** The diagram shows a cross-sectional view of the tire with the inner and outer radii labeled. The outer radius extends to 35 cm, and the inner radius is 30 cm. --- ### Explanation of Diagram The diagram displays a circular representation of the tire's cross-section: 1. **Outer Radius (r_outer)**: This is the distance from the center of the tire to the outer edge, labeled as 35 cm. 2. **Inner Radius (r_inner)**: This is the distance from the center of the tire to the inner edge, labeled as 30 cm. The calculation of the force of friction will consider the average radius when the brakes apply force, which is halfway between the inner and outer radius. --- ### Steps to Calculate the Magnitude of Force of Friction: 1. **Moment of Inertia (I):** \[ I_{\text{tire}} = \frac{1}{2} M (r_{\text{outer}}^2 + r_{\text{inner}}^2) \] Given: - \( M = 1.8 \) kg - \( r_{\text{outer}} = 0.35 \) m - \( r_{\text{inner}} = 0.30 \)
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