Roger Waters’ Latest Rant: He Isn’t ‘Interested’ in Eddie Van Halen or AC/DC
British rock icon Roger Waters has been in some, well, hot water lately with controversial comments about a range of socio-political topics.
Among them, the Ukraine war and "lies" of Russian war crimes, the United States' involvement aiding Ukraine in the conflict and also placing culpability on American and British Jewish people for the actions of Israel, as seen in the Pink Floyd founding member's latest explosive interview with Rolling Stone on October 4.
Waters' comments have been so inflammatory that a series of his solo This Is Not A Drill shows in Poland for April 2023 were canceled and Variety has reported that the backlash has put the prospective sale of Pink Floyd's music catalog (for an estimated $500 million) in jeopardy.
Though not nearly as damaging as his global affair comments, Waters' latest opinions might strike a chord with music fans. In a new three-hour interview with "The Joe Rogan Experience," the musician shared that he's not a fan of "loud" rock 'n' roll or legends Eddie Van Halen or AC/DC.
"[I'm] not really interested in loud rock 'n' roll — which some people are and they love it, but I couldn't care less about AC/DC or Eddie Van Halen or any of that stuff. It's just, who? I don't go, 'Who?' because obviously I know the name. And I'm sure Eddie's brilliant and a great guitar player and wonderful. It just doesn't interest me."
Waters also admitted that he's "not very up on rock history," adding, "I'm not very interested in most popular music. I mean, there are certain people that I'm great fans of, mainly the writers, the singer-songwriters, you know. So, [Bob] Dylan and Neil Young. But I won't start a long list, because I probably could, but it's that end of the spectrum that I'm more interested in."
The musician also revealed to Rogan that he's working on a book with revelations about Pink Floyd, calling the material "hard things to write about." Given his latest word spewing, we're sure he won't be holding back.