b. A 4.0-cm-diameter, 55 g ball is shot horizontally into a tank of 40°C honey. How long will it take for the horizontal speed to decrease to 10% of its initial value?

Question

Accepted Answer

Hey, everyone. So this problem is dealing with drag forces. Let's see what they're asking us. We have a bead moving in the X direction in an oil medium of viscosity given to us as a equals 45.4 times 10 to the negative three pascal seconds. The beads radius is two millimeters and its mass is g at time. T equals zero seconds. The velocity of the bead is 25 cm/s. We are asked to calculate the time required for the bead to reach a velocity of 12.5 cm/s. And we are told to neglect the gravitational force. So the first thing, oh So in our multiple choice answers here are a Q A 0.81 seconds. B 2.0 seconds, C 6.1 seconds or D 36. seconds. OK. So the first thing we can do here is draw our free body diagram of the bead. We are told to neglect the gravitational force. So the only force acting on this speed is our drag force and is acting in the negative X direction because the drag force is negative to the um direction of movement. And so we are, we have a sphere bead, we are working in a viscous liquid. Those are two hints that we can use Stokes Law. So we can recall that Stokes Law is given as the drag force equals six pi times eight times, sorry, six pi multiplied by a multiplied by R multiplied by V. And for Newton's second law, we can recall that the sum of the forces in the X direction is equal to mass multiplied by acceleration. And we've already established that the only force in the X direction acting on the speed is our drag force. So we can rewrite this as negative F D equals mass multiplied by acceleration. So they are asking us to calculate a time and neither of these equations have time in them. But the next thing we can do is recall that acceleration is a, is the derivative of velocity with respect to time. So we can rewrite acceleration as DVD T. So this equation becomes negative drag force equals the mass multiplied by DVD T. And we can set the drag forces equal to each other and rearrange all of this. So we'll have a negative six pi beta R V equals M DVD T. Now, when we're working with, in a girls, we wanna get, we want to get the um terms the variables and the derivatives on the same side. So we will rearrange this equation to have our D V and V on the same side. So that will look like D V divided by B equals negative six pi A R divided by M D team. OK. We can integrate, integrate both sides. And so we'll integrate D V over V from V I to V F. For this, the right hand side of the equation, we can pull out the constants. So that's negative six pi beta R and M R divided by M. And then we will have simply the integral of D T from T equals zero seconds to T. The integral of D V over V is L N of the, from VI 2 VF. And then that equals negative six pi beta R divided by M time multiplied by T from time 0 to 10. We can recall from our log map that the L N of V F minus the L N of V I can be rewritten as the L N of V F divided by V I. And that's going to make this equation easier to solve. So that is how we are going to uh solve this next step. So we have the L N of V F divided by V I equals -6 PA to R T divided by M. So now we have tea and that is what we are solving for here. We can rearrange to isolate that variable T equals L N of V F divided by V I multiplied by M all divided by -6 Pi Beta R. And from here, we can go back up into the problem and make sure we have all of these variables, write them all out and then plug in chug to solve for T. So the problem gave us a final velocity of 12.5 cm/s and an initial velocity of 25 cm/s. Now, because we are dividing these two um values by each one by one another. I'm not going to worry about putting those units into Standard form because they'll just cancel out. OK. Next, we have a mass that was given to us as 15 g. So that will be 15 times 10 to the negative three kg. Keeping that in standard form is important here or standard units six pi and then a our viscosity was given to us as 45. times 10 to the negative three pascal seconds. And R the radius of our bead was given as two millimeters which equals two times 10 to the negative three meters. And we were right, we have everything we needed to plug in. So our time equals the L N of 12.5 centimeters per second, divided by 25 cm/s. Multiplied by our mass times 10 to the negative 3 kg divided by our died by negative six pi multiplied by 45.4 times 10 to the -3 pass seconds multiplied by R. Radius 2 times 10 to the negative 3 m from there we can solve for time. We get 6.1 seconds. So that is the correct answer for this problem. When we look at our multiple choice solutions that aligns with answer C so that's all we have for this one. We'll see you in the next video.

b. A 4.0-cm-diameter, 55 g ball is shot horizontally into a tank of 40°C honey. How long will it take for the horizontal speed to decrease to 10% of its initial value?

Verified Solution

Key Concepts

Drag Force

Newton's Second Law of Motion

Viscosity