An iron tyre of diameter 50cm at 15 degree celsius is to be shrunk on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm?(co-efficient of linear expansion of iron 0.0000121 degree celsius)
Well, let's see if we can iron out this problem, shall we? 🤡
To find the temperature at which the tyre will slip over the wheel, we can use the principle of thermal expansion. The change in diameter is related to the change in temperature by the coefficient of linear expansion. So, let's break it down step by step:
1. First, let's calculate the change in diameter of the tyre:
ΔD = 50.35 cm - 50 cm = 0.35 cm
2. Next, let's convert the radial gap into centimeters:
Radial gap = 0.5 mm = 0.05 cm
3. Now, we can calculate the change in diameter required to accommodate the radial gap:
ΔD_required = ΔD + 2 * Radial gap = 0.35 cm + 2 * 0.05 cm = 0.45 cm
4. Finally, let's calculate the change in temperature needed using the coefficient of linear expansion:
ΔT = ΔD_required / (Coefficient of linear expansion)
= 0.45 cm / (0.0000121 °C^(-1))
≈ 3727.27 °C
Wow, that's quite a hot temperature! You might need some serious firepower to get that tyre to slip over the wheel. But don't worry, I doubt you'll ever need to heat up your tyre that much. Just make sure to keep a safe distance, and don't try this at home! 🌡️🚗💥
To calculate the temperature to which the iron tyre must be heated in order to slip over the wheel with a radial gap of 0.5 mm, we can use the concept of thermal expansion.
1. Find the change in diameter of the iron tyre:
Δd = (π * 50.35 cm) - (π * 50 cm)
= 0.35π cm
2. Since the gap is radial, we need to find the change in radius:
Δr = Δd / 2
= (0.35π cm) / 2
= 0.175π cm
3. Convert the change in radius to meters:
Δr_m = 0.175π cm * (1 m / 100 cm)
= 0.175π m
4. Use the formula for linear expansion:
ΔL = α * L0 * ΔT
where:
ΔL is the change in length
α is the coefficient of linear expansion
L0 is the original length
ΔT is the change in temperature
In this case, we need to use the diameter instead of length:
Δd = α * d0 * ΔT
Rearrange the formula to solve for ΔT:
ΔT = Δd / (α * d0)
Substitute the given values:
ΔT = (0.175π m) / (0.0000121 1/°C * 50 cm)
5. Calculate ΔT:
ΔT = (0.175π m) / (0.0000121 1/°C * 0.5 m)
≈ 45,714.88 °C
Therefore, the iron tyre must be heated to approximately 45,714.88 °C in order to slip over the wheel with a radial gap of 0.5 mm.
To find the temperature at which the iron tyre must be heated in order to slip over the wheel with a radial gap of 0.5mm, we can use the concept of linear expansion.
The linear expansion of a material is given by the formula:
ΔL = (α * L * ΔT)
Where:
ΔL is the change in length,
α is the coefficient of linear expansion,
L is the original length or diameter, and
ΔT is the change in temperature.
In this case, we have:
ΔL = 0.5mm = 0.05cm
α = 0.0000121/°C (coefficient of linear expansion of iron)
L = diameter of the iron tyre = 50cm
ΔT = ?
We need to find the change in temperature (ΔT) when the iron tyre is heated. Rearranging the formula, we have:
ΔT = ΔL / (α * L)
Substituting the values, we get:
ΔT = 0.05cm / (0.0000121/°C * 50cm)
Simplifying the equation, we find:
ΔT ≈ 82.644.63°C
Therefore, the iron tyre must be heated to approximately 82.64°C in order for it to slip over the wheel with a radial gap of 0.5mm.
R wheel = 50.35/2 + 0.005cm = 25.18 cm
Rtire= 25
so
d Rtire / R tire = 0.18/50 = coef * d Temp