# Express the force of the rockets thrust which has a

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express the force of the rocket’s thrust, which has a magnitude of T 30N, in terms of the unit vectors, ~ F thrust T cos 30 ˆ r T sin 30 ˆ θ. The rod will act to negate the force of the rocket in ˆ r direction, so the total force on the rocket, ~ F , is ~ F T sin 30 ˆ θ 30 sin 30 ˆ θ 15N ˆ θ. By Newton’s second law, the sum of the forces on the ˆ θ direction is equal to the mass times the acceleration in the ˆ θ direction of the object. 9

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F θ ma θ 15N . We are given the mass of the object, so we can find the tangential accel- eration by simply dividing the tangential force by the objects mass. a θ 15N 2kg 7 . 5 meters second 2 . Because the object will move in a circle, we can convert this linear accel- eration into an angular acceleration by dividing by the radius of the circle the motion will trace out. α a θ R 7 . 5 meters second 2 5 meters radius 1 . 5 1 radii sec 2 1 . 5 radians second 2 . Because the tangential acceleration was a constant, and the angular ac- celeration is simply the tangential acceleration scaled by a constant, we know that the angular acceleration is also a constant, this allows us to invoke the angular kinematics equations. The question asks us for the angular veloc- ity of the object after it has completed two turns, so we must first use the equation θ f θ 0 ω 0 t 1 2 αt 2 to solve for the time it takes to make two complete rotations, then use the equation ω f ω 0 αt to solve for the final velocity after that time has elapsed. We must be careful to match units, we have α in terms of rad sec 2 , so we must be sure to express θ in radians. The distance traveled if one were to walk along the circumference of a circle is given by C 2 πR . To convert this to radians, we must divide by the radius of the circle. So we have that C R 2 πR R 2 π which very well may be a familiar relation to many students. Then, the distance traveled by traversing the circle twice will be twice as many radians, thus we have that 10
1Rev 2 π rad 2Rev 2 2 π rad 4 π rad . Then inserting 4 π radians into the angular kinematics equations, we get that: 4 π 0 0 t 1 2 1 . 5 t 2 4 π 3 4 t 2 t 16 π 3 4 . 1 sec. This is the time it takes to traverse the circle twice, finding the angular velocity at this time is equivalent to finding the angular velocity when the object has completed two revolutions, so inserting t 4 . 1 seconds into the other angular kinematic equation: ω f ω 0 αt ω f 0 1 . 5 4 . 1 ω f 6 . 15 radians second , which is closest to answer a. 11

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7. A metal bead is thrown at and angle θ above the horizontal with initial speed v 0 inside a viscous liquid. The magnitude of drag force ~ D is well modelled by ~ D b v 2 x v 2 y , where b is a constant and ~v v x , v y is the instantaneous velocity of the bead with respect to the liquid (in a frame where the x -axis points horizontally and the y -axis points vertically up). The differential equations that best describe the motion of the bead are (a) dv x dt b m v x , dv y dt g b m v y 52 (b) dv x dt b m v 2 x v 2 y , dv y dt g b m v 2 x v 2 y (c) dv x dt b m v x v , dv y dt g b m v y v (d) dv x dt b m
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• Fall '07
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