QuestionDecember 20, 2025

Suppose you exert a force of 212 N tangential to a 0235 m radius 75.0 kg grindstone (a solid disk). (Give your answers to at least 2 decimal places.) (a) What torque (inNcdot m) is exerted? (Enter the magnitude.) square N-m (b) What is the angular acceleration (inrad/s^2) assuming negligible opposing friction?(Enter the magnitude.) square rad/s^2 (c) What is the angular acceleration (i) nrad/s^2) if there is an opposing frictional force of 20.6 N exerted 1.25 cm from the axis?(Enter the magnitude.) square rad/s^2

Suppose you exert a force of 212 N tangential to a 0235 m radius 75.0 kg grindstone (a solid disk). (Give your answers to at least 2 decimal places.) (a) What torque (inNcdot m) is exerted? (Enter the magnitude.) square N-m (b) What is the angular acceleration (inrad/s^2) assuming negligible opposing friction?(Enter the magnitude.) square rad/s^2 (c) What is the angular acceleration (i) nrad/s^2) if there is an opposing frictional force of 20.6 N exerted 1.25 cm from the axis?(Enter the magnitude.) square rad/s^2
Suppose you exert a force of 212 N tangential to a 0235 m radius 75.0 kg grindstone (a solid disk). (Give your answers to at least 2 decimal places.) (a) What torque (inNcdot m) is exerted? (Enter the magnitude.) square N-m (b) What is the angular acceleration (inrad/s^2) assuming negligible opposing friction?(Enter the magnitude.) square rad/s^2 (c) What is the angular acceleration (i) nrad/s^2) if there is an opposing frictional force of 20.6 N exerted 1.25 cm from the axis?(Enter the magnitude.) square rad/s^2

Solution
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Answer

(a) 49.82\,\text{N}\cdot\text{m} ### (b) 25.74\,\text{rad/s}^2 ### (c) 24.40\,\text{rad/s}^2 Explanation 1. Calculate Torque (a) Torque: \tau = rF. Here, r = 0.235\,\text{m}, F = 212\,\text{N}. So, \tau = 0.235 \times 212. 2. Find Moment of Inertia For a solid disk: I = \frac{1}{2}MR^2. M = 75.0\,\text{kg}, R = 0.235\,\text{m}. 3. Calculate Angular Acceleration without Friction (b) \alpha = \frac{\tau}{I}. 4. Calculate Net Torque with Friction (c) Friction torque: \tau_f = r_f F_f, r_f = 0.0125\,\text{m}, F_f = 20.6\,\text{N}. Net torque: \tau_{net} = \tau - \tau_f. 5. Angular Acceleration with Friction \alpha_{fric} = \frac{\tau_{net}}{I}.

Explanation

1. Calculate Torque (a)<br /> Torque: $\tau = rF$. Here, $r = 0.235\,\text{m}$, $F = 212\,\text{N}$. So, $\tau = 0.235 \times 212$.<br />2. Find Moment of Inertia<br /> For a solid disk: $I = \frac{1}{2}MR^2$. $M = 75.0\,\text{kg}$, $R = 0.235\,\text{m}$.<br />3. Calculate Angular Acceleration without Friction (b)<br /> $\alpha = \frac{\tau}{I}$.<br />4. Calculate Net Torque with Friction (c)<br /> Friction torque: $\tau_f = r_f F_f$, $r_f = 0.0125\,\text{m}$, $F_f = 20.6\,\text{N}$. Net torque: $\tau_{net} = \tau - \tau_f$.<br />5. Angular Acceleration with Friction<br /> $\alpha_{fric} = \frac{\tau_{net}}{I}$.
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