An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min, calculate:

Question

Asked on August 3, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-08-03T04:28:18.071Z

Here are the solutions to the questions. QUESTION 1: GENERAL 1.1 Define momentum. Momentum is a measure of the mass in motion of an object. It is defined as the product of an object's mass and its velocity. 1.2 Distinguish between angular velocity and angular acceleration. Angular velocity* () is the rate of change of angular displacement, indicating how fast an object rotates or revolves about an axis. It is measured in radians per second (rad/s). Angular acceleration* () is the rate of change of angular velocity, indicating how quickly the angular velocity of an object is changing. It is measured in radians per second squared (rad/s^2). 1.3 State Hooke's law. Hooke's Law states that the force (F) needed to extend or compress a spring by some distance (x) is directly proportional to that distance. Mathematically, F = kx, where k is the spring constant. 1.4 Distinguish between elasticity and limit of proportionality. Elasticity* is the property of a material that allows it to return to its original shape and size after deforming forces are removed. The limit of proportionality* is the point on a stress-strain curve beyond which stress is no longer directly proportional to strain. Up to this limit, Hooke's Law is obeyed. QUESTION 2: KINEMATICS 2.1 Calculate the velocity of vehicle A relative to vehicle B. Step 1: Resolve the velocities of vehicle A and vehicle B into their x and y components. Vehicle A: v_A = 105 km/h northwest (angle _A = 135^ from positive x-axis). v_Ax = 105 (135^) ≈ -74.246 km/h v_Ay = 105 (135^) ≈ 74.246 km/h Vehicle B: v_B = 115 km/h directly east (angle _B = 0^ from positive x-axis). v_Bx = 115 (0^) = 115 km/h v_By = 115 (0^) = 0 km/h Step 2: Calculate the components of the relative velocity v_A/B = v_A - v_B. v_A/Bx = v_Ax - v_Bx = -74.246 - 115 = -189.246 km/h v_A/By = v_Ay - v_By = 74.246 - 0 = 74.246 km/h Step 3: Calculate the magnitude of the relative velocity. |v_A/B| = sqrt((-189.246)^2 + (74.246)^2) = sqrt(35893.9 + 5512.4) = sqrt(41406.3) ≈ 203.485 km/h Step 4: Calculate the direction of the relative velocity. = ((74.246)/(-189.246)) ≈ -21.42^ Since the x-component is negative and the y-component is positive, the angle is in the second quadrant. _actual = 180^ - 21.42^ = 158.58^ The velocity of vehicle A relative to vehicle B is 203.5 km/h at 158.6^. 2.2 Calculate the resulting velocity in magnitude and direction. Step 1: Resolve the ship's velocity and current's velocity into their x and y components. Ship velocity: v_s = 95 km/h south-east (angle _s = -45^ from positive x-axis). v_sx = 95 (-45^) ≈ 67.175 km/h v_sy = 95 (-45^) ≈ -67.175 km/h Current velocity: v_c = 20 km/h E 25° N (angle _c = 25^ from positive x-axis). v_cx = 20 (25^) ≈ 18.126 km/h v_cy = 20 (25^) ≈ 8.452 km/h Step 2: Calculate the components of the resulting velocity v_R = v_s + v_c. v_Rx = v_sx + v_cx = 67.175 + 18.126 = 85.301 km/h v_Ry = v_sy + v_cy = -67.175 + 8.452 = -58.723 km/h Step 3: Calculate the magnitude of the resulting velocity. |v_R| = sqrt((85.301)^2 + (-58.723)^2) = sqrt(7276.26 + 3448.39) = sqrt(10724.65) ≈ 103.56 km/h Step 4: Calculate the direction of the resulting velocity. = ((-58.723)/(85.301)) ≈ -34.54^ The resulting velocity is 103.6 km/h at -34.5^ (or E 34.5^ S). 2.3 A boy throws a cricket ball at an angle of 20° to the horizontal with an initial velocity of 45 m/s. Step 1: Resolve the initial velocity into horizontal and vertical components. u_x = 45 (20^) ≈ 42.2865 m/s u_y = 45 (20^) ≈ 15.390 m/s The acceleration due to gravity is a_y = -9.81 m/s^2. 2.3.1 The time it will take to reach maximum height. At maximum height, the vertical velocity v_y = 0. v_y = u_y + a_y t 0 = 15.390 m/s + (-9.81 m/s^2) t t = (15.390)/(9.81) ≈ 1.5688 s The time to reach maximum height is 1.57 s. 2.3.2 The maximum height reached by the ball. Using the vertical motion equation: v_y^2 = u_y^2 + 2 a_y y 0^2 = (15.390 m/s)^2 + 2 (-9.81 m/s^2) y 0 = 236.8521 - 19.62 y y = (236.8521)/(19.62) ≈ 12.072 m The maximum height reached by the ball is 12.07 m. 2.3.3 The horizontal displacement of the ball. The total time of flight is twice the time to reach maximum height: T = 2 × 1.5688 s = 3.1376 s. The horizontal displacement is given by x = u_x T (since there is no horizontal acceleration). x = 42.2865 m/s × 3.1376 s ≈ 132.61 m The horizontal displacement of the ball is 132.6 m. QUESTION 3: ANGULAR MOTION 3.1 An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min, calculate: Step 1: Convert given values to SI units. Diameter D = 11 cm = 0.11 m, so radius R = D/2 = 0.055 m. Time t = 6 s. Initial angular velocity _0 = 0 rad/s (starts from rest). Final angular speed N = 1380 r/min. Convert to rad/s: _f = 1380 revmin × 2 rad1 rev × 1 min60 s = (1380 × 2)/(60) rad/s = 46 rad/s ≈ 144.513 rad/s 3.1.1 The angular acceleration of the pulley in rad/s^2. Using the angular kinematic equation: _f = _0 + t 144.513 rad/s = 0 rad/s + (6 s) = 144.513 rad/s6 s ≈ 24.0855 rad/s^2 The angular acceleration of the pulley is 24.09 rad/s^2. 3.1.2 The linear acceleration of the vee belt on the pulley in m/s^2. The linear acceleration a is related to angular acceleration by a = R. a = 24.0855 rad/s^2 × 0.055 m ≈ 1.3247 m/s^2 The linear acceleration of the vee belt is 1.32 m/s^2. 3.1.3 The maximum linear velocity of the vee belt in m/s. The maximum linear velocity v is related to the final angular velocity _f by v = _f R. v = 144.513 rad/s × 0.055 m ≈ 7.9482 m/s The maximum linear velocity of the vee belt is 7.95 m/s. Drop the next question.

Accepted Answer · 2026-08-03T04:28:18.071Z

Here are the solutions to the questions. QUESTION 1: GENERAL 1.1 Define momentum. Momentum is a measure of the mass in motion of an object. It is defined as the product of an object's mass and its velocity. 1.2 Distinguish between angular velocity and angular acceleration. Angular velocity* () is the rate of change of angular displacement, indicating how fast an object rotates or revolves about an axis. It is measured in radians per second (rad/s). Angular acceleration* () is the rate of change of angular velocity, indicating how quickly the angular velocity of an object is changing. It is measured in radians per second squared (rad/s^2). 1.3 State Hooke's law. Hooke's Law states that the force (F) needed to extend or compress a spring by some distance (x) is directly proportional to that distance. Mathematically, F = kx, where k is the spring constant. 1.4 Distinguish between elasticity and limit of proportionality. Elasticity* is the property of a material that allows it to return to its original shape and size after deforming forces are removed. The limit of proportionality* is the point on a stress-strain curve beyond which stress is no longer directly proportional to strain. Up to this limit, Hooke's Law is obeyed. QUESTION 2: KINEMATICS 2.1 Calculate the velocity of vehicle A relative to vehicle B. Step 1: Resolve the velocities of vehicle A and vehicle B into their x and y components. Vehicle A: v_A = 105 km/h northwest (angle _A = 135^ from positive x-axis). v_Ax = 105 (135^) ≈ -74.246 km/h v_Ay = 105 (135^) ≈ 74.246 km/h Vehicle B: v_B = 115 km/h directly east (angle _B = 0^ from positive x-axis). v_Bx = 115 (0^) = 115 km/h v_By = 115 (0^) = 0 km/h Step 2: Calculate the components of the relative velocity v_A/B = v_A - v_B. v_A/Bx = v_Ax - v_Bx = -74.246 - 115 = -189.246 km/h v_A/By = v_Ay - v_By = 74.246 - 0 = 74.246 km/h Step 3: Calculate the magnitude of the relative velocity. |v_A/B| = sqrt((-189.246)^2 + (74.246)^2) = sqrt(35893.9 + 5512.4) = sqrt(41406.3) ≈ 203.485 km/h Step 4: Calculate the direction of the relative velocity. = ((74.246)/(-189.246)) ≈ -21.42^ Since the x-component is negative and the y-component is positive, the angle is in the second quadrant. _actual = 180^ - 21.42^ = 158.58^ The velocity of vehicle A relative to vehicle B is 203.5 km/h at 158.6^. 2.2 Calculate the resulting velocity in magnitude and direction. Step 1: Resolve the ship's velocity and current's velocity into their x and y components. Ship velocity: v_s = 95 km/h south-east (angle _s = -45^ from positive x-axis). v_sx = 95 (-45^) ≈ 67.175 km/h v_sy = 95 (-45^) ≈ -67.175 km/h Current velocity: v_c = 20 km/h E 25° N (angle _c = 25^ from positive x-axis). v_cx = 20 (25^) ≈ 18.126 km/h v_cy = 20 (25^) ≈ 8.452 km/h Step 2: Calculate the components of the resulting velocity v_R = v_s + v_c. v_Rx = v_sx + v_cx = 67.175 + 18.126 = 85.301 km/h v_Ry = v_sy + v_cy = -67.175 + 8.452 = -58.723 km/h Step 3: Calculate the magnitude of the resulting velocity. |v_R| = sqrt((85.301)^2 + (-58.723)^2) = sqrt(7276.26 + 3448.39) = sqrt(10724.65) ≈ 103.56 km/h Step 4: Calculate the direction of the resulting velocity. = ((-58.723)/(85.301)) ≈ -34.54^ The resulting velocity is 103.6 km/h at -34.5^ (or E 34.5^ S). 2.3 A boy throws a cricket ball at an angle of 20° to the horizontal with an initial velocity of 45 m/s. Step 1: Resolve the initial velocity into horizontal and vertical components. u_x = 45 (20^) ≈ 42.2865 m/s u_y = 45 (20^) ≈ 15.390 m/s The acceleration due to gravity is a_y = -9.81 m/s^2. 2.3.1 The time it will take to reach maximum height. At maximum height, the vertical velocity v_y = 0. v_y = u_y + a_y t 0 = 15.390 m/s + (-9.81 m/s^2) t t = (15.390)/(9.81) ≈ 1.5688 s The time to reach maximum height is 1.57 s. 2.3.2 The maximum height reached by the ball. Using the vertical motion equation: v_y^2 = u_y^2 + 2 a_y y 0^2 = (15.390 m/s)^2 + 2 (-9.81 m/s^2) y 0 = 236.8521 - 19.62 y y = (236.8521)/(19.62) ≈ 12.072 m The maximum height reached by the ball is 12.07 m. 2.3.3 The horizontal displacement of the ball. The total time of flight is twice the time to reach maximum height: T = 2 × 1.5688 s = 3.1376 s. The horizontal displacement is given by x = u_x T (since there is no horizontal acceleration). x = 42.2865 m/s × 3.1376 s ≈ 132.61 m The horizontal displacement of the ball is 132.6 m. QUESTION 3: ANGULAR MOTION 3.1 An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min, calculate: Step 1: Convert given values to SI units. Diameter D = 11 cm = 0.11 m, so radius R = D/2 = 0.055 m. Time t = 6 s. Initial angular velocity _0 = 0 rad/s (starts from rest). Final angular speed N = 1380 r/min. Convert to rad/s: _f = 1380 revmin × 2 rad1 rev × 1 min60 s = (1380 × 2)/(60) rad/s = 46 rad/s ≈ 144.513 rad/s 3.1.1 The angular acceleration of the pulley in rad/s^2. Using the angular kinematic equation: _f = _0 + t 144.513 rad/s = 0 rad/s + (6 s) = 144.513 rad/s6 s ≈ 24.0855 rad/s^2 The angular acceleration of the pulley is 24.09 rad/s^2. 3.1.2 The linear acceleration of the vee belt on the pulley in m/s^2. The linear acceleration a is related to angular acceleration by a = R. a = 24.0855 rad/s^2 × 0.055 m ≈ 1.3247 m/s^2 The linear acceleration of the vee belt is 1.32 m/s^2. 3.1.3 The maximum linear velocity of the vee belt in m/s. The maximum linear velocity v is related to the final angular velocity _f by v = _f R. v = 144.513 rad/s × 0.055 m ≈ 7.9482 m/s The maximum linear velocity of the vee belt is 7.95 m/s. Drop the next question.

An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min, calculate:

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An electric motor has a vee pulley with an effective diameter of 11 cm on its shaft. If the motor, after being switched on, takes 6 seconds to reach maximum speed of 1380 r/min, calculate:

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