A uniform solid cylinder of mass 1kg rolls with out slipping on aflate surface at aspeed of pay/200rad/s how long will it be when they meet?
To find how long it will take for two objects to meet, we need to analyze their relative motion and the distance between them.
In this case, we have a uniform solid cylinder rolling without slipping on a flat surface. Let's assume the other object is stationary.
First, let's calculate the linear speed of the cylinder. The linear speed of a rolling object is given by the equation:
v = ω * r
where v is the linear speed, ω is the angular speed (in radians per second), and r is the radius of the cylinder.
You mentioned the angular speed as "pay/200rad/s," but it seems there might be a typographical error. Assuming that you meant "π/200 rad/s" instead of "pay/200 rad/s" (where π represents the mathematical constant Pi, approximately equal to 3.14159), we can continue with the calculation.
Given:
ω = π/200 rad/s
r = radius of the cylinder
Next, we need to determine the time taken for the cylinder to meet the stationary object. To do that, we need to calculate the distance covered by the rolling cylinder before they meet.
The distance covered by the rolling cylinder is equal to the circumference of the cylinder. The circumference can be calculated using the formula:
C = 2πr
Now, let's assume the time taken to meet is t. The linear distance covered by the rolling cylinder during time t is given by:
d = v * t
Since the rolling cylinder and the stationary object meet at the same time, the linear distance covered by both objects is the same. Hence, we have:
d = C
Combining the equations and solving for time:
2πr = v * t
Since we have the relationship between linear speed and angular speed:
v = ω * r
Now we can substitute this into the equation:
2πr = ω * r * t
Canceling out "r" on both sides:
2π = ω * t
Finally, solving for t:
t = 2π / ω
Now, plug in the given angular speed:
t = 2π / (π/200) = 2 * 200 = 400 seconds
Therefore, it will take 400 seconds (or 6 minutes and 40 seconds) for the cylinder to meet the stationary object.