A PERSON LOWERS A BUCKET INTO A WELL BY TURNING THE HAND CRANK. THE CRANK HANDLE MOVES WITH A CONSTANT TANGENTIAL SPEED OF 120M/S ON ITS CIRCULAR PATH. THE ROPE HOLDING THE BUCKET UNWINDS WITH OUT SLIPPING ON THE BARREL OF THE CRANK. FIND THE LINEAR SPEED WITH WHICH THE BUCKET MOVES DOWN THE WELL
120 = speed of bucket * radius of handle/radius of barrel
To find the linear speed with which the bucket moves down the well, we need to understand the relationship between the tangential speed of the hand crank and the linear speed of the bucket.
The tangential speed of the hand crank is given as 120 m/s. This means that the distance covered by a point on the circumference of the crank handle in one second is 120 meters.
We can use this tangential speed to find the linear speed of the bucket by equating the linear speed of the bucket with the tangential speed of the hand crank.
Let's assume the radius of the circular path of the crank handle is 'r' meters. We need to determine the linear speed of the bucket, which we'll call 'v'.
The linear speed of the bucket can be calculated using the formula:
v = ω * r
where ω is the angular velocity of the crank handle, and 'r' is the radius of the circular path.
In this case, since the tangential speed of the hand crank is given, we can find the angular velocity using the formula:
ω = v / r
By substituting the given tangential speed of the hand crank into the formula, we have:
ω = 120 m/s / r
Now we can substitute this angular velocity back into the formula for the linear speed of the bucket:
v = ω * r
v = (120 m/s / r) * r
Simplifying the equation:
v = 120 m/s
Therefore, the linear speed with which the bucket moves down the well is 120 m/s.