A 350g air track cart on a horizontal air track is attached toa string that goes

Question

A 350g air track cart on a horizontal air track is attached toa string that goes over a pulley with a moment of inertia of 6.00x10-6 kg*m2 and a radius of 1.35 cm.The string is pulled vertically downward by a force of 2.5 N. Whatis the tension in the string between the pulley and the cart? Also. When a car is weight it is allowed to move over a scalewhich records a reading as the front wheels go over the scale and asecond reading when the rear wheels go over the scale. The weightof the cair is equal to the weight under the front wheels the weight under the rear wheels the average of the two weights the sum of the two weights the difference of the two weights A 350g air track cart on a horizontal air track is attached toa string that goes over a pulley with a moment of inertia of 6.00x10-6 kg*m2 and a radius of 1.35 cm.The string is pulled vertically downward by a force of 2.5 N. Whatis the tension in the string between the pulley and the cart? Also. When a car is weight it is allowed to move over a scalewhich records a reading as the front wheels go over the scale and asecond reading when the rear wheels go over the scale. The weightof the cair is equal to the weight under the front wheels the weight under the rear wheels the average of the two weights the sum of the two weights the difference of the two weights

Explanation / Answer

the mass of the air track is m = 350 g = 350 * 10-3kg the moment of inertia of the pulley is I = 6.00 *10-6 kg.m2 the radius of the pulley is r = 1.35 cm = 1.35 *10-2 m The string is pulled vertically downward by a force of F = 2.5N the troque acting on the string is = I * ---------(1) is the angular acceleration of the pulley the torque on the string is also given by = F * r or = 1.5 * 1.35 * 10-2 = 2.025 *10-2 N-m therefore from (1) we get = (/I) = (2.025 * 10-2/6.00 *10-6) = 3375 rad/s2 the tangential acceleration of the string is aT = * r = 3375 * 1.35 *10-2 = 45.56 m/s2 therefore,the tension in the string between the pulley and thecart is T = m * aT = 350 * 10-3 * 45.56 = 15.9N let the weight of the car be w when the front wheel goes over the scale then the torqueacting on the car is 1 = w * (-d) when the rear wheel goes over the scale then the torqueacting on the car is 2 = w * (d) the net torque acting on the car is net = (2 -1) the net torque acting on the car is equal to the work done bythe car therefore the force acting on the car is F = (net/d - (-d)) =(net/2d) or F = (2 - 1/2d) = ((w * d- (w * (-d))/2d) = (2wd/2d) = w the force acting on the car is equal to the weight of thecar.Therefore,the weight of the car is equal to the difference ofthe two weights. therefore,the tension in the string between the pulley and thecart is T = m * aT = 350 * 10-3 * 45.56 = 15.9N let the weight of the car be w when the front wheel goes over the scale then the torqueacting on the car is 1 = w * (-d) when the rear wheel goes over the scale then the torqueacting on the car is 2 = w * (d) the net torque acting on the car is net = (2 -1) the net torque acting on the car is equal to the work done bythe car therefore the force acting on the car is F = (net/d - (-d)) =(net/2d) or F = (2 - 1/2d) = ((w * d- (w * (-d))/2d) = (2wd/2d) = w the force acting on the car is equal to the weight of thecar.Therefore,the weight of the car is equal to the difference ofthe two weights. when the rear wheel goes over the scale then the torqueacting on the car is 2 = w * (d) the net torque acting on the car is net = (2 -1) the net torque acting on the car is equal to the work done bythe car therefore the force acting on the car is F = (net/d - (-d)) =(net/2d) or F = (2 - 1/2d) = ((w * d- (w * (-d))/2d) = (2wd/2d) = w the force acting on the car is equal to the weight of thecar.Therefore,the weight of the car is equal to the difference ofthe two weights.

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