# Given the current length of the day is 24 hrs we can

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Given the current length of the day is 24 hrs we can find ω now , the angular velocity of the earth at its current rotation rate, is ω now 1Rev Day 2 π radians 1Rev 1Day 24hours 1hour 3600seconds 2 π radians 24 3600 seconds We are told that the length of the day increases by 1%, so the new angular velocity of the earth ω new is given by ω new 2 π radians 1 . 01 24 3600 seconds . The problem asks for the new difference between the poles in terms of the old difference. Inserting ω now into the equation for Δ a we find that Δ a now ω 2 now R 2 π radians 24 3600 seconds 2 R . Similarly, by inserting ω new into the equation for Δ a we find that: Δ a new ω 2 new R 2 π radians 1 . 01 24 3600 seconds 2 R 1 1 . 01 2 Δ a now . 98Δ a now . Thus Δ a has decreased by 2%. 5

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3. A car is turning left as it goes into a curve in the road. A rider of mass m experiences a fictitious force of magnitude F pulling him right, out of the curve. In the same car, at the same time, a second rider of mass 2 m experiences a fictitious force out of the curve of mangintude (a) 2 F . 52 (b) F . (c) 1 2 F . (d) 1 4 F . (e) 4 F . When a car is turning it is undergoing an acceleration, thus it is a non- inertial reference frame and we must again consider the centrifugal force. The equation for the centrifugal force is F c 2 r. Because both riders are in the same car and undergoing a turn of the same radius, we know that both ω and r are the same for both riders. Therefore, if the mass of the first rider is M then they experience a force of F 1 2 r . Then, given the mass of the second rider is 2 M , we have that F 2 2 2 r 2 F 1 . 4. 6
2 m m v v Two blocks slide on a frictionless straight track towards each other as shown in the figure. After they collide the total momen- tum (a) points right. 52 (b) points left. (c) vanishes. (d) cannot be determined unless we know whether the two carts stick after the collision. (e) has changed by the impulse J in the collision. This problem is about the conservation of momentum. If we call the ˆ x direction along the short side of the page with positive ˆ x to the right we have that the initial momentum of the entire system, ~ p 0 , is the sum of the momenta of each object in the system before the collision. ~ p 0 ~ p m ~ p 2 m mv ˆ x 2 mv ˆ x mv ˆ x. Because the initial and final momentum of the system must be equal, we know that ~ p f ~ p 0 mv ˆ x . Thus the total momentum after the collision points in the ˆ x direction, which we defined as to the right. 5. A bomb of mass 8 m at rest explodes into three fragments. The first one, of mass 3 m , is ejected horizontally with speed v . The second fragment, of mass 4 m , is also ejected horizontally and with speed v but in a direction perpendicular to the first one. The third fragment is ejected ... (a) with speed 5 v . 52 (b) with speed 7 v .

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• Fall '07
• Hicks

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