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A wheel of radius r rolls without slippi...

A wheel of radius `r` rolls without slipping with a speed `v` on a horizontal road. When it is at a point `A` on the road, a small blob of mud separates from'the wheel at- its highest point and lands at point `B` on the road:

A

`AB = v sqrt((r)/(g))`

B

`AB = 2v sqrt((r)/(g))`

C

`AB = 4v sqrt((r)/(g))`

D

If `v gt sqrt(4rg)`, the blob mud will land on the wheel and not on the road

Text Solution

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The correct Answer is:
To solve the problem of finding the distance between points A and B when a blob of mud separates from a rolling wheel, we can follow these steps: ### Step-by-Step Solution 1. **Understand the Scenario**: - A wheel of radius \( r \) rolls without slipping on a horizontal road with a speed \( v \). - At point A (the point of contact with the road), a blob of mud separates from the highest point of the wheel and lands at point B. 2. **Identify the Initial Conditions**: - The wheel is rolling with a linear speed \( v \). - The height from which the blob of mud falls is equal to the diameter of the wheel, which is \( 2r \). 3. **Determine the Time of Flight**: - The blob of mud will fall under the influence of gravity. The vertical displacement \( s \) is given by: \[ s = -2r \] - The initial vertical velocity \( u_y = 0 \) (since it starts falling from rest). - The acceleration \( a_y = -g \) (acceleration due to gravity). - Using the equation of motion: \[ s = ut + \frac{1}{2} a t^2 \] - Substituting the known values: \[ -2r = 0 \cdot t + \frac{1}{2} (-g) t^2 \] - This simplifies to: \[ -2r = -\frac{1}{2} g t^2 \implies t^2 = \frac{4r}{g} \implies t = 2\sqrt{\frac{r}{g}} \] 4. **Calculate the Horizontal Distance**: - The horizontal distance \( x \) that the blob travels while falling can be calculated using the horizontal velocity. - The horizontal velocity \( u_x \) at the point of separation is equal to the linear speed of the wheel, which is \( v \). - The horizontal distance \( x \) is given by: \[ x = u_x \cdot t \] - Substituting the values: \[ x = v \cdot \left(2\sqrt{\frac{r}{g}}\right) = 2v\sqrt{\frac{r}{g}} \] 5. **Final Result**: - Therefore, the distance between points A and B is: \[ x = 4v\sqrt{\frac{r}{g}} \]
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