Question 15.14: The crane shown rotates with a constant angular velocity V1 ......
The crane shown rotates with a constant angular velocity \text V_1 of 0.30 rad/s. Simultaneously, the boom is being raised with a constant angular velocity \text V_2 of 0.50 rad/s relative to the cab. Knowing that the length of the boom OP is l = 12 m, determine (a) the velocity of the tip of the boom, (b) the acceleration of the tip of the boom.
Frames of Reference. The frame OXYZ is fixed. We attach the rotating frame Oxyz to the cab. Its angular velocity with respect to the frame OXYZ is therefore Ω = \text V_1 = (0.30 rad/s)j. The angular velocity of the boom relative to the cab and the rotating frame Oxyz (or F for short) is \text V_{B/F}=\text V_2 = (0.50 rad/s)k.
a. Velocity v_P. From Eq. (15.46) we write
\text v _P=\text v _{P^{\prime}}+\text v _{P / F} (1)
where \text v _{P^{\prime}} is the velocity of the point P^{\prime} of the rotating frame which coincides with P:
\text v _{P^{\prime}}= \Omega \times r =(0.30~ rad / s ) j \times[(10.39~ m ) i +(6~ m ) j ]=-(3.12~ m / s ) kand where \text v _{P/F} is the velocity of P relative to the rotating frame Oxyz. But the angular velocity of the boom relative to Oxyz was found to be \text V_{B/F} = (0.50 rad/s)k. The velocity of its tip P relative to Oxyz is therefore
Substituting the values obtained for \mathrm{v}_{P^{\prime}} \mathrm{~and~} \mathrm{v}_{P /{F}} into (1), we find
\mathrm{v}_{P}=-(3 \mathrm{~m} / \mathrm{s}) {i}+(5.20 \mathrm{~m} / \mathrm{s}) {j}-(3.12 \mathrm{~m} / \mathrm{s}) {k}b. Acceleration aP. From Eq. (15.48) we write
{a}_{P}={a}_{P^{\prime}}+{a}_{P /{F}}+{a}_{c} (2)
Since Ω and \text {V}_{B / F} are both constant, we have
Substituting for {a}_{P^{\prime}}, {a}_{P / {F}},\mathrm{~and}~ {a}_{c} into (2), we find
{a}_{P}=-\left(3.54 \mathrm{~m} / \mathrm{s}^{2}\right) {i}-\left(1.50 \mathrm{~m} / \mathrm{s}^{2}\right) {j}+\left(1.80 \mathrm{~m} / \mathrm{s}^{2}\right) {k}