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Chapter 17: Problem 87

The \(100-\mathrm{kg}\) pendulum has a center of mass at \(G\) and a radius of gyration about \(G\) of \(k_{G}=250 \mathrm{~mm}\). Determine the horizontal and vertical components of reaction on the beam by the pin \(A\) and the normal reaction of the roller \(B\) at the instant \(\theta=90^{\circ}\) when the pendulum is rotating at \(\omega=8 \mathrm{rad} / \mathrm{s}\). Neglect the weight of the beam and the support.

Short Answer

Expert verified
The vertical component of force at A (pin) is 981 N downwards, the horizontal component of force at A (pin) is 1600 N towards right, and the normal reaction at B (roller) is 490.5 N upwards.

Step by step solution

01

Determine the Angular Momentum

To determine the reactions, first calculate the moment of inertia \(I_G\) of the pendulum about the point \(G\). The moment of inertia \(I_G\) equals to \(m*k_G^2\), where \(m\) is the mass of the pendulum and \(k_G\) is the radius of gyration about \(G\). Substituting the given values, we obtain: \(I_G = 100 kg * (0.25 m)^2 = 6.25 kg*m^2\). Next, calculate the angular momentum \(L\), which is \(I*\omega\). Substituting the values of \(I_G\) and \(\omega=8 rad/s\), we obtain: \(L = 6.25 kg*m^2 * 8 rad/s = 50 kg*m^2/s\).
02

Calculate the Centripetal Force and Torque

Next, calculate the centripetal force using the formula: \(F_c=m*\omega^2* r\), where \(r\) is the length of the pendulum. Since the length of pendulum is equal to radius of gyration \(k_G = 0.25 m\), the centripetal force \(F_c \) is: \(F_c = 100 kg * (8 rad/s)^2 * 0.25 m = 1600 N\). We can then calculate the torque about point A using the relationship \(\tau = F_c * r * sin(90)\), which yields: \(\tau = 1600 N * 0.25 m * 1 = 400 N*m\).
03

Calculate Reactions at A and B

Afterward, calculate the vertical and horizontal components of force at A. Vertical component of force at A (\(R_{Ay}\)) is equal to the weight of pendulum (\(W\)) and can be computed as: \(R_{Ay} = m*g = 100 kg*9.81 m/s^2 = 981 N\). The horizontal component of force at A (\(R_{Ax}\)) can be found using the equation of motion along x-axis: \(R_{Ax} = F_c = 1600 N\), that implies that the pin at A induced a force of 1600 N horizontally to provide the required centripetal force. Finally, calculate the normal reaction at B (\(R_B\)) from the condition of rotational equilibrium about A which is: If \(\sum \tau_A = 0\), then \(R_B * l - W * l/2 = 0\), which gives: \(R_B = W/2 = 490.5 N\).

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Most popular questions from this chapter

The 20 -kg roll of paper has a radius of gyration \(k_{A}=90 \mathrm{~mm}\) about an axis passing through point \(A\). It is pin supported at both ends by two brackets \(A B\). If the roll rests against a wall for which the coefficient of kinetic friction is \(\mu_{k}=0.2\), determine the constant vertical force \(F\) that must be applied to the roll to pull off \(1 \mathrm{~m}\) of paper in \(t=3 \mathrm{~s}\) starting from rest. Neglect the mass of paper that is removed.

The \(20-\mathrm{kg}\) punching bag has a radius of gyration about its center of mass \(G\) of \(k_{G}=0.4 \mathrm{~m}\). If it is initially at rest and is subjected to a horizontal force \(F=30 \mathrm{~N}\), determine the initial angular acceleration of the bag and the tension in the supporting cable \(A B\).

The uniform disk of mass \(m\) is rotating with an angular velocity of \(\omega_{0}\) when it is placed on the floor. Determine the time before it starts to roll without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is \(\mu_{k}\).

The disk of mass \(m\) and radius \(r\) rolls without slipping on the circular path. Determine the normal force which the path exerts on the disk and the disk's angular acceleration if at the instant shown the disk has an angular velocity of \(\boldsymbol{\omega}\).

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