Let\'s use Newton\'s second law for rotation to find the acceleration of a bucke
ID: 1409282 • Letter: L
Question
Let's use Newton's second law for rotation to find the acceleration of a bucket (mass m) in an old-fashioned well, and the angular acceleration of the winch cylinder. SOLUTION SET UP We start by sketching the situation, using the information from (Figure 1). The winch cylinder has mass M and radius R the bucket has mass m and falls a distance h. We must treat the two objects separately, so we draw a free-body diagram for each (Figure 2). We take the positive direction of the y coordinate for the bucket to be downward; the positive sense of rotation for the winch is counterclockwise. We assume that the rope is massless and that it unwinds without slipping. SOLVE Newton's second law applied to the bucket (mass m) gives sigma F_g = mg - T = ma_y The forces on the cylinder are its weight Mg, the upward normal force n exerted by the axle, and the rope tension T. The first two forces act along lines through the axis of rotation and thus have no torque with respect to that axis. The moment arm of the force T on the cylinder is R, and the torque is tau = TR. Applying sigma tau = l_alpha and the expression for the moment of inertia of a solid cylinder to the winch cylinder, we find that sigma Tau = l_alpha and TR = 1/2 MR^2 alpha The angular acceleration a of the cylinder is proportional to the magnitude of acceleration a_y of the unwinding rope: a_y = R_alpha. We use this equation to replace (R_alpha) with a_y in the cylinder equation, and then we divide by R. The result is T = 1/2 Ma_y Now we substitute this expression for T into the equation of motion for the bucket (mass m): mg - 1/2 Ma_y = ma_y When we solve this equation for a_y, we get a_y = g/1+M/2m Combining this result with the kinematic equation a_y = R_alpha, we obtain the angular acceleration alpha: alpha = g/R/1+M/2m Finally, we can substitute the equation for a_y back into the equation sigma F vector = ma vector for m to get an expression for the tension T in the rope: T = mg - ma_y = mg - m g/1+M/2m and T = mg/1+2m/M REFLECT First, note that the tension in the rope is less than the weight mg of the bucket; if the two forces were equal (in magnitude), the bucket wouldn't accelerate downward. We can check two particular cases. When M is much larger than m (a massive winch and a little bucket), the ratio m/M is much smaller than unity. Then the tension is nearly equal to mg, and the acceleration a_y is correspondingly much less than g. When M is zero, T = 0 and a = g; the bucket then falls freely. If it starts from rest at a height h above the water, its speed nu when it strikes the water is given by nu^2 = squareroot 2a_yh =squareroot 2gh/1+M/2m thus, What is the tension T in the rope if the bucket's acceleration is half the acceleration of free fall?Explanation / Answer
for the bucket force equation is given as
mg - T = ma
given a = g/2
so mg - T = mg/2
T = mg/2
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